= Solution
The <anticommutator> condition and invertibility give the <matrix similarities>
$$
B^{-1}AB=-A,\qquad A^{-1}BA=-B.
$$
Since <matrix trace> is invariant under <matrix similarity>,
$$
\operatorname{Tr}A=\operatorname{Tr}(-A)=-\operatorname{Tr}A,\qquad\operatorname{Tr}B=-\operatorname{Tr}B.
$$
Hence $\boxed{\operatorname{Tr}A=\operatorname{Tr}B=0}$. Taking <determinants> in the first <matrix similarity> also gives
$$
\det A=\det(-A)=(-1)^n\det A.
$$
Since $A$ is an <invertible matrix>, $\det A\ne0$, 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.
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