 Solution

The anticommutator condition and invertibility give the matrix similaritiesSince matrix trace is invariant under matrix similarity,Hence $\boxed{\mathrm{Tr}A=\mathrm{Tr}B=0}$. Taking determinants in the first matrix similarity also givesSince $A$ is an invertible matrix, $\det A\ne 0$, and $\boxed{n\text{ is even}}$. In terms of eigenvalues, the same matrix similarity pairs each nonzero eigenvalue $\lambda$ with $-\lambda$, with the same algebraic multiplicity, so an odd number of them is impossible.