If I understand correctly, the argument you're making is this. You know that in situations A and B, the fields between the top and bottom capacitors aren't going to be the same: that's what is meant by "current must flow". The charges will redistribute, because you know what the steady state in situation A is, and you know what the steady state in situation B is, and so you know that there must be some kind of redistribution. This is what quantumwave's answer is saying.
However, the paradox is that if you could freeze the charges in place, and move the top capacitor plates closer together, then as we go from situation A to B, the electric field acting on the unequal charges in the top capacitor plates hasn't changed, and neither has it changed on the charges in the bottom plates, and so there is no force to pull extra charges from the bottom to the top plates.
I can think of three possible arguments to explain what's happening here.
* In order for the two plates to be brought closer together, you need to move them. This means they will have a non-zero velocity as we move from situations A to B. 531
Therefore, there will be magnetic effects, and you can't use the results for the electrostatic steady state distribution of the charges on the capacitor plates. Although you can make this velocity $v$ as small as you like (quasi-static), the non-static field effects will be of the same order of magnitude as $v$, and so these could cause redistribution of the charges at a rate of order $v$, and if you integrate the non-static charge redistribution over the whole path, it will sum up to the required change of charge.
I find the above argument not so convincing, as in theory, we could fix the electrical charges in place during the transition from A to B. For example, we could cut the wire, then bring the top plates closer together, and then reconnect the wire afterwards. In other words, there must still be a path from the non-equilibrium situation B to the steady state.
* The more convincing argument is that because of the introduction of the conducting wire between the bottom plates and the top plates, the charges in the wire and at the connection between the wire and the plates are not at equilibrium, so they will start moving, which will "kick-start" the motion of the charges towards equilibrium.
* Finally, another line of argument might be that it's caused by the edge effects of the capacitor.
I would like to be give more details, by explaining exactly how the wires modify the electric field, but it's probably going to be quite complicated. I thought it might be interesting to study a toy model, where the plates are all parallel: you put down a capacitor at $z = 0$, then another at $z = z_0$, connect the plates by a vertical wire at $x = y = 0$, and then to connect the topmost plate to the bottommost plate, you make the whole of space loop along the $z$ axis, either by repeating the pattern, or by modifying the topology of the space. Unfortunately I don't have time to work through it, but if somebody else would like to give it some thought, that would be great. 780