The title of the question is misleading.
In the body of the question we read-
"My question is purely about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft."
Obviously for acceleration to be zero, net force must be zero, so net vertical force must be equal to weight. This is not a very interesting question.
The title asks a completely different question: "Does lift equal weight in a climb?" This is a much more interesting question.
In the context of fixed-winged flight, Lift is defined to act perpendicular to the flight path through the airmass, and Drag is defined to act parallel to the flight path through the airmass. For the purposes of the answer, we'll assume that Thrust acts parallel to the flight path through the airmass, although this clearly not always exactly true. This simplifying assumption leads to the following vector diagram:
Powered climb at climb angles of 45 and 90 degrees:
In the vector diagrams above, "angle c" is the climb angle-- it is 45 degrees in the left-hand figure, and 90 degrees in the right-hand figure.
We can see that in a powered climb, Lift = Weight * cosine (climb angle), where the climb angle is measured relative to the airmass (an important distinction in the case of gliding flight-- an unpowered climb in a thermal updraft is still a descent in relation to the airmass!)
Clearly, Lift is less than Weight in a powered climb. For example, if the climb angle is 45 degrees, Lift = .707 * Weight. If the climb angle is 90 degrees, Lift must be zero.
The same is also true in a descent-- Lift = Weight * cosine (descent angle), so Lift is less than Weight. This is explored in more detail in some of the links given at the end of this answer.
Note that we've taken the approach of combining the Thrust and Drag vectors into a single (Thrust-Drag) vector, and then we've arranged this vector into a closed vector triangle with Lift and Weight. Whenever vectors can be arranged nose-to-tail into into a closed polygon-- a triangle in this case-- this shows that net force must be zero, meaning that acceleration is zero and velocity i constant. For clarity we've also drawn the individual Thrust and Drag vectors outside the vector triangle. These are redundant with the (Thrust-Drag) vector.
Varying the climb angle and/or the L/D ratio:
Note that for a given aircraft in a given configuration, any given angle-of-attack is associated with specific values for the lift coefficient, drag coefficient, and ratio of lift coefficient / drag coefficient. Lift is proportional to lift coefficient * airspeed squared, and drag is proportional to drag coefficient * airspeed squared, so the ratio of the lift coefficient / drag coefficient is also the ratio of Lift / Drag. So for a given aircraft in a given configuration, any given angle-of-attack is associated with specific ratio of Lift to Drag.
If the left-hand diagram above and the middle diagram above both represent the same aircraft in the same configuration, then the aircraft must be flying slightly slower in the middle diagram. That's the only way the L and D values can both be slightly smaller, for the same L/D ratio. Adding power to increase the climb angle, while holding the angle-of-attack constant, makes the airspeed decrease slightly. However in the case illustrated here, the change in airspeed would be too tiny to ever notice in practice-- it would be equal to the square root of the change in the value of the magnitude of the lift vector or drag vector.
If all the diagrams represent the same aircraft in the same configuration of flaps etc, then the right-hand diagram (5:1 L/D ratio) would represent a lower angle-of-attack than the left-hand or middle diagrams (10:1 L/D ratio). (We'll ignore the other possibility that the 5:1 case represents mushing flight very near stall, where drag is very high.) A lower angle-of-attack means a lower lift coefficient, yet the size of the lift vector is the same, so the airspeed must be higher in the case illustrated in the right-hand diagram. Therefore the climb rate is also higher. In short, when we increase thrust to increase our climb rate, we also must reduce angle-of-attack, if for some reason we wish to keep our climb angle constant rather than allowing it to increase.
Powered climb at 45-degree climb angle at 8 different ratios of Lift to Drag:
Note that as we decrease our L/D ratio, it takes more and more thrust to maintain the same 45-degree climb angle. In the case where the L/D ratio is 2/1, thrust must actually be greater than weight! This is a bit counterintuitive, as we could obviously climb straight up with some small but non-zero airspeed if thrust were only slightly greater than weight. However, that vertical climb would be conducted at a very low airspeed. In the diagram above, if all cases represent the same aircraft in the same configuration, by constraining the climb angle to be constant, so that L also must remain constant, we're constraining the airspeed to get progressively higher and higher as we reduce the angle-of-attack, lift coefficient, and L/D ratio. Hence the huge increase in drag, and thrust-required, as we reduce the angle-of-attack, lift coefficient, and L/D ratio.
As we explore climb angles closer and closer to 90 degrees, the L/D ratio has less and less influence on the thrust required. A figure similar to the one above, but for a climb angle of 60 or 70 degrees, would show less increase n thrust-required as we decrease the angle-of-attack, lift coefficient, and L/D ratio than we see at a climb angle of 45 degrees. This also implies that we're forcing less of an increase in airspeed as we decrease the angle-of-attack, lift coefficient, and L/D ratio in such a case. That makes sense-- as thrust carries more and more of the aircraft weight, the dynamics of the wing have less and less influence upon airspeed. In the case of a truly vertical climb, the wing must be at the zero-lift angle-of-attack and the L/D ratio must be zero. In such a case, of course, the drag force still varies with airspeed, and so the faster we want to fly straight up, the more thrust we need.
For the sake of clarity, this answer has focussed on some rather steep climb angles. It's important to also keep in mind that for shallow climb angles (or descent) angles that are typical of general aviation light aircraft, the cosine of the climb angle is not much smaller than 1, and so Lift is nearly equal to Weight (specifically, Lift is only slightly less than Weight.) Since Weight doesn't vary with climb or dive angle, we can conclude that for shallow climb or dive angles-- with no other accelerations going on (specifically, the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn)-- Lift is also nearly constant, regardless of whether the aircraft is climbing, descending, or neither. This means if the climb or descent angle is shallow and the net G-load is one, the airspeed indicator can also be interpreted as an angle-of-attack gauge. Why should this be so? To keep lift nearly constant, it must be approximately true that the lift coefficient is varying in inverse proportion to the square of the airspeed. This establishes a nearly fixed relationship between airspeed and angle-of-attack, for shallow climb or descent angles and net G-loadings near one. If the airspeed is low, the lift coefficient and angle-of-attack must be high, and if the airspeed is high, the lift coefficient and angle-of-attack must be low, regardless of whether the aircraft is climbing at a shallow angle, descending at a shallow angle, or flying horizontally. So the airspeed indicator is in essence an angle-of-attack gauge at shallow climb or descent angles. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- a given angle-of-attack will be associated with a lower airspeed than in horizontal flight, and a given airspeed will be associated with a lower angle-of-attack than in horizontal flight. In the most extreme case where the aircraft is climbing straight up, lift must be zero, so the lift coefficient must be zero, and the angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads. Clearly the airspeed indicator cannot serve "double duty" as a guide to angle-of-attack in such a situation.
We've also assumed throughout this answer that the Thrust vector acts parallel to the flight path through the airmass. Obviously, if this isn't true, then the equation lift = weight * cosine (climb angle) is also no longer true. To take an extreme case, note that when the exhaust nozzles of a Harrier "jump jet" are pointed straight down, the wing is "unloaded"-- the plane can hover at zero airspeed with zero lift, supported entirely by thrust. Conversely, during a glider winch launch, the towline pulls steeply downward on the glider. This too can be viewed as a form of "vectored thrust"-- but now the load on the wing is increased, rather than decreased, so the wings must generate a lift force that is much greater than the aircraft's weight. At any rate, it's best to thoroughly understand the simple case where the thrust vector acts parallel to the flight path, before going on to consider more exotic cases.
To see a vector diagram of the forces in climbing flight from an outside reference source, see the diagram below. This diagram shows the same relationships as the other diagrams included in this answer, but the forces have not been arranged into a closed vector polygon, so it is less obvious that the net force is zero.
Above is a vector diagram showing the forces in a stabilized, linear, constant-airspeed climb -- from https://systemdesign.ch/wiki/L%C3%B6sung_zu_Steigflug
FS = thrust
FW = drag
FGp is the component of weight that acts parallel to the flight path, and is ALSO exactly equal in magnitude and opposite in direction to (thrust - drag)
FGs is the component of weight that acts perpendicular to the flight path, and is ALSO exactly equal in magnitude and opposite in direction to lift.
FA = lift
FG = weight
Angle beta is the climb angle-- the angle between the flight path and the horizon.
Another answer to the present question addresses a "climbing by increasing airspeed" case, which is also characterized by a completely fixed pitched attitude. Perhaps a better description of this case is "entering a climb without pitching up at all". This introduces a downthrust component relative to the flight path. This downthrust is the root cause of the increase in the lift vector above and beyond the "standard" value of weight * cosine (climb angle).
The present answer assumes no downthrust or upthrust. If downthrust or upthrust is present, the full equation for the magnitude of the lift vector is weight * cosine (climb angle) + thrust * sine (downthrust angle), where the downthrust angle is measured relative to the flight path, not the horizon. Treat upthrust as negative downthrust.
When we start in level cruise with no downthrust, and then enter a climb without pitching up at all, the downthrust angle is now so large that lift is indeed greater than weight. The climb angle and the downthrust angle are the same in this situation. Note that as we increase thrust and begin climbing without allowing the aircraft to pitch up at all, the wing's angle-of-attack must decrease, which likely means that we are pushing forward on the control yoke or stick. Naturally, this is not the normal way to conduct a climb! See this ASE answer for more on the "climbing without allowing the aircraft to pitch up" case, in the context of an another analogous situation -- the "propeller train".
See these related answers to related questions:
"Are there any situations where having high lift but low lift to drag ratio would be beneficial?"
"What produces Thrust along the line of flight in a glider?"
"'Gravitational' power vs. engine power"
"Descending on a given glide slope (e.g. ILS) at a given airspeed— is the size of the lift vector different in headwind versus tailwind?"
"Are we changing the angle of attack by changing the pitch of an aircraft?"
"Is excess lift or excess power needed for a climb?"