What is the maximum number of elements of the inverse of an $n\times n$ matrix consisting of natural numbers $(\geq1)$ that are identical to itself?
This question arose from my previous question, How many elements in the inverse of an $n\times n$ positive matrix can be positive? . There is an upper bound $n^2-n$ because at least $n$ elements must be negative, as I wrote in that question. Also, I found a series of matrices such that $n^2-2n+2$ as shown below. I think it can be shown by the product of the matrix and its inverse being the identity matrix. So, the maximum should be between $n^2-2n+2$ and $n^2-n$.
$n=2$ $$\begin{pmatrix}2&3\\1&2\end{pmatrix}^{-1}=\begin{pmatrix}\color{red}2&-3\\-1&\color{red}2\end{pmatrix}$$ $n=3$ $$\begin{pmatrix}3&4&4\\1&2&1\\1&1&2\end{pmatrix}^{-1}=\begin{pmatrix}\color{red}3&-4&-4\\-1&\color{red}2&\color{red}1\\-1&\color{red}1&\color{red}2\end{pmatrix}$$ $n=4$ $$\begin{pmatrix}4&5&5&5\\1&2&1&1\\1&1&2&1\\1&1&1&2\end{pmatrix}^{-1}=\begin{pmatrix}\color{red}4&-5&-5&-5\\-1&\color{red}2&\color{red}1&\color{red}1\\-1&\color{red}1&\color{red}2&\color{red}1\\-1&\color{red}1&\color{red}1&\color{red}2\end{pmatrix}$$ $n=5$ $$\begin{pmatrix}5&6&6&6&6\\1&2&1&1&1\\1&1&2&1&1\\1&1&1&2&1\\1&1&1&1&2\end{pmatrix}^{-1}=\begin{pmatrix}\color{red}5&-6&-6&-6&-6\\-1&\color{red}2&\color{red}1&\color{red}1&\color{red}1\\-1&\color{red}1&\color{red}2&\color{red}1&\color{red}1\\-1&\color{red}1&\color{red}1&\color{red}2&\color{red}1\\-1&\color{red}1&\color{red}1&\color{red}1&\color{red}2\end{pmatrix}$$
Edit: I changed the direction of the 2 in the above matrices. And here is the proof of this. $$\left(\begin{array}{c|ccc}n&&\mathbf{(n+1)}_{n-1}&\\\hline&&&\\\mathbf{1}_{n-1}&&J_{n-1}+I_{n-1}&\\&&&\end{array}\right)\left(\begin{array}{c|ccc}n&&\mathbf{(-n-1)}_{n-1}&\\\hline&&&\\\mathbf{-1}_{n-1}&&J_{n-1}+I_{n-1}\\&&&\end{array}\right)=\left(\begin{array}{c|ccc}n^2+(\mathbf{n+1})\cdot(-\mathbf{1})&&n(\mathbf{-n-1})+(\mathbf{n+1})J+(\mathbf{n+1})&\\\hline&&&\\n\mathbf{1}+(J+I)(\mathbf{-1})&&1(\mathbf{-n-1})+J^2+JI+IJ+I&\\&&&\end{array}\right)\\=\left(\begin{array}{c|ccc}n^2-(n+1)\cdot(n-1)&&\left((-n^2-n)+(n+1)(n-1)+(n+1)\right)\mathbf{1}&\\\hline&&&\\\left(n-(n-1)-1\right)\mathbf{1}&&(-n-1)J+(n-1)J+2J+I&\\&&&\end{array}\right)\\=\left(\begin{array}{c|ccc}1&&\mathbf{0}&\\\hline&&&\\\mathbf{0}&&I&\\&&&\end{array}\right)=I$$
where $J$ is a unit matrix, consisting of all 1s.
