Hi,
One can apply even-odd mode analysis to understand the operation of Willkinson divider.
Even mode: the middle point of resistor is open circuit and resistor is not contained in the circuit, so the even part is indeed losless.
Odd mode: the middle point of resistor is short circuited so the resitor is contained in the circuit and contributes to the losses.
For the generall case of the Willkinson divider operation, these two should be added so for the generall case of two circuits connected to the outputs there will be loss in the resistor that will contribute to the isolation between the ports.
For the particular case when the signal is split into two identical matched circuits (2x50Ohm termination, and accordingly nothing else!) there are no reflected signals, the circuit exhibits perfect symmetry and odd mode cannot exist at all, so all of the ports are isolated, so indeed it can be lossless for this particular case.
However, this is rearly ever the case of using the real Willkinson divider. When the terminations are mutually equal (symmetrical) but not matched, no reflected power will be disipated on the resistor (again, it is perfect symmetry) but it will appear on the input port.
If the terminations are not matched and are not mutually equal, part of the reflected power (imbalance, odd mode) will be spent in the resistor and part (balanced, even mode) reflected to the input.
When doing the acctual calculation about even/odd mode part quantities by hand, one must take into account that S parameters are complex quantities so the terminations must be considered as complex impedances too (phase of the reflected signals is of course very important). Any simulator does this automatically, so one just have to interpret the results)
flyhigh