A contradiction regarding flux conservation
Your contradiction is based on your false model of the physics:

In steady state, the voltage across L1 and L2 is zero, each, simply because a inductor has zero effective resistance to DC.

So, closing the switch ironically won't do anything in this idealized model: with zero voltage across the switch, there's no reason for current to flow through the switch.

So, closing the switch changes nothing in this idealized model.

One could, however, just as well say: well, if I put two perfect conductors (the closed switch and L2 for DC) in parallel, then the current would be split half-and-half; but that's really just a fallacy, because you simply can't find a reason that this should be happening. The current-divider rules you know only apply to cases where there's some impedance in both branches; that's not the case here, and the math becomes ill-defined.

What will happen with this circuit in reality is that of course, an inductor has ohmic resistance, and so the voltage across L2 isn't actually zero. A current will hence flow through the switch once closed, in the opposite direction as the current through L2. That current will need to come from "somewhere". So, yes, the current through L2 will change in reality, but certainly not SUDDENly, to put it in your words. See Scott's answer for a discussion of the differential equation you'd need to solve.