Finding matrices with $A_1^{-1}+A_2^{-1}+...+A_k^{-1}=(A_1+A_2+...+A_k)^{-1}$

If $k$ is odd, then it is easy: take $A_{2m} = -A_{2m+1}$ for every $m\ge 1$ so that the whole expression reduces to $A_1^{-1} = A_1^{-1}$ that is true for every $A_1$ invertible.

If $k$ is even, then using the same trick as before, what's left is $$ A_1^{-1} + A_2^{-1} = (A_1+A_2)^{-1} $$ So if you solve it for $k=2$, you solve it for every $k$.

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Let's prove that $n=3$ and $k=2$ produces an absurd. It turns out that $$ 2I + A_1A_2^{-1} + A_2A_1^{-1} = I $$ and if $X = A_1A_2^{-1}$ then $X^{-1} = A_2A_1^{-1}$ and $$ X^2 + X + I = 0. $$ This means that the minimal polynomial of $X$ is $x^2+x+1$, so the characteristic polynomial must be $(x^2+x+1)^m$ for some $m$, but then $n=2m=3$, that is impossible.