<s>[INST] Problem of existing matrices 
Let $n,k \geq 2$ be two integers. Prove that there must exist some $n \times n$ invertible, non-diagonal matrices: $X_1,X_2, \dots , X_k$ with real entries, such that $$X_1^{-1}+X_2^{-1}+\dots+X_k^{-1}=(X_1+X_2+\dots+X_k)^{-1}$$

Since this is a problem which asks to construct the matrices, I tried to make it less general. I tried to find some matrices with $X_i^2=I_n$ and hence $X_i^{-1}=X_i$, so what would be left to take care of is the equality $(X_1+X_2+\dots+X_k)^2=I_n.$ 
For the case when $k$ is odd, I chose $X_1$ to be the matrix with $1$ on the antidiagonal and $0$ anywhere else; then for $X_i=(-1)^iX_1,\: i \geq 2$ the problem is solved.
However, I have problems when $k$ is even. I couldn't find such matrices even when $k=2$.
 [/INST] Here's a method that uses complex numbers. Let us note the association 
\begin{align}
a+ib \longleftrightarrow
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}.
\end{align}
Hence it suffices to consider solutions to 
\begin{align}
\frac{1}{z_1+z_2} = \frac{1}{z_1} + \frac{1}{z_2}  \implies -1 = \frac{z_1}{z_2} + \frac{z_2}{z_1} = z + \frac{1}{z}.  
\end{align}
In  particular, let us focus on solutions lying on the unit circle. Hence we get
\begin{align}
-1 = e^{i\theta} + e^{-i\theta} = 2\cos \theta 
\end{align}
which means $\theta = \frac{2\pi}{3}$ is a solution, that is,
\begin{align}
\frac{z_1}{z_2} = e^{i\frac{2\pi}{3}} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}. 
\end{align}
Let us choose $z_1 = i$ and $z_2 = \frac{\sqrt{3}}{2}-\frac{i}{2}$, then the corresponding matrices 
\begin{align}
X_1 = 
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\ \ \ \text{ and } \ \ 
X_2 = 
\begin{pmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2}\\
-\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix}
\end{align}
satisfy the relation
\begin{align}
(X_1+X_2)^{-1} = X_1^{-1} + X_2^{-1}.
\end{align}
</s>