It seems a bit silly and I guess I know the answer but I don't know why it gave me some confusion, in a half-bridge configuration like the one in the image:
If between DC-Bus+ and DC-Bus- there is a voltage of 400VDC (for example), the voltage that exists between each drain and source of the MOSFETs when they are OFF, is it half of DC-Bus+? that is, Vds(1) = Vds(2) = V(DC-Bus+) / 2 if the two mosfets are of the same model?
It depends.
If the bridge is switching, and you're talking about the deadtime between ON times then maybe not. Current in the inductor can slew the switch node to just below ground or just above the positive rail (depending on the current and direction in the inductor) causing a diode drop higher than the DC bus voltage to appear across one of the FETs.
If you're talking about just turning off both FETs with no load and perfectly matched FETs, in steady-state, then you could reasonably expect half of the DC bus to appear across each FET.
If between DC-Bus+ and DC-Bus- there is a voltage of 400VDC (for example), the voltage that exists between each drain and source of the MOSFETs when they are OFF, is it half of DC-Bus+?
Well, that's irrelevant to your title question: -
How much voltage must a MOSFET withstand in a half-bridge configuration?
When one MOSFET is "on", the "off" MOSFET will see the full DC bus voltage of 400 volts and, under these circumstances, I'd be looking to use a MOSFET that is rated at 500 volts or above.
