Why does Kramer's theorem only apply to systems with an odd number of fermions? Kramer's theorem states that: In a time reversal symmetric system, where $T^2=-1$, all energy levels are (at least) doubly degenerate.
I claim to have proven this in a simple way.
We first establish that $|\Psi\rangle$ and $T|\Psi\rangle$ are different states. Let's assume the opposite, that the states are equivalent, then, $$T|\Psi\rangle=e^{i\phi}|\Psi\rangle.$$
Which implies that, $$T^2|\Psi\rangle=Te^{i\phi}|\Psi\rangle=e^{-i\phi}T|\Psi\rangle=e^{-i\phi+i\phi}|\Psi\rangle=|\Psi\rangle.$$
From which it follows that $T^2=1$, contrary to our requirement. Thus $|\Psi\rangle$ and $T|\Psi\rangle$ are unique states. Then trivially, $H|\Psi\rangle=E|\Psi\rangle$ and $$HT|\Psi\rangle=TH|\Psi\rangle=ET|\Psi\rangle.$$ So that both $|\Psi\rangle$ and $T|\Psi\rangle$ have the same energy! The system is doubly degenerate.
But nowehere was it assumed that the system must have an odd number of fermions. What am I missing here?
 A: You are assuming $T^2=-1$. You can have $T^2=1$ too, in which case it is perfectly fine to have $\psi$ and $T\psi$ be the same state, as you showed in your question. But it turns out the states with $T^2=-1$ are just the states with half-integer angular momentum, as I'll show.
$T$ anticommutes with angular momentum (since it should flip the sign). If I consider a particle at rest, and use a basis in terms of eigenstates of z-angular momentum $\psi_\sigma$, we have
$$T\psi_\sigma = \zeta_\sigma \psi_{-\sigma}$$
where $\zeta_\sigma$ is some phase factor that might depend on $\sigma$. Notice we see right away that we can't have $T\psi\propto\psi$ for a particle with half integer spin, since there is no $\sigma=0$ state. That is more or less what you wanted to see I think but let's go a little further.
If we apply the raising and lowering operator $J_\pm=J_x\pm iJ_y$
$$J_\pm T \psi_\sigma=\zeta_\sigma J_\pm \psi_{-\sigma}\propto \zeta_\sigma \psi_{-\sigma\pm 1}$$
where there's also that normalization factor with a square root you can look up. But we also have,
$$J_{\pm}T\psi_\sigma = -TJ_\mp \psi_\sigma \propto -\zeta_{\sigma\mp 1}\psi_{-\sigma\pm 1} $$
The normalization factor is the same in both cases so
$$\zeta_{\sigma}=-\zeta_{\sigma\mp 1}$$
So the sign flips each time you raise or lower z-angular momentum, so we can write (up to an overall phase we can absorb into our definition of $\psi$)
$$\zeta_\sigma = (-1)^{j-\sigma}$$
where $j$ is the total angular momenta. So we have
$$T^2\psi_\sigma=(-1)^{j+\sigma}(-1)^{j-\sigma}\psi_\sigma=(-1)^{2j}\psi_\sigma$$
So $T^2=\pm 1$ depending on whether $j$ is an integer or half integer.
