Let $A$ be an $n \times n$ matrix with entries $1, 2, \dots, n^2$ (without repeating), e.g., when $n=2$,
$$ A \in \left\{ \begin{bmatrix} 1& 2\\ 3&4 \end{bmatrix}, \ \ \begin{bmatrix} 1 & 4 \\ 2 & 3\end{bmatrix}, \dots \right\} $$
Let $r(A)$ denote the rank of $A$. What are the possible values of $r(A)$?
Clearly, $r(A) \in \{ 1, 2, \dots, n\}$ and I have constructed rank-$2$ and rank-$n$ matrices. I guess $r(A)$ can be any natural number from $2$ to $n$, but I don't know how to construct matrix with rank $k$, where $2<k<n$.
I suspect that for any $2 < k \leq n$, the following procedure results in a matrix containing entries $1,2,\dots,n^2$ has rank $k$. I've checked it for all $3 \leq n \leq 20$ and $2 < k \leq n$; I hope to add a proof once I have the time.
For example, with $n = 6$ and $k = 4$ we get the matrix $$ A = \begin{pmatrix}\color{red}2 & \color{red}1 & 3 & 4 & 5 & 6\\ 7 & \color{red}9 & \color{red}8 & 10 & 11 & 12\\13 & 14 & 15 & 16 & 17 & 18\\19 & 20 & 21 & 22 & 23 & 24\\25 & 26 & 27 & 28 & 29 & 30\\31 & 32 & 33 & 34 & 35 & 36\end{pmatrix}, $$ which indeed has rank $4$.
So far, I am able to show that the statement holds for $2<k<n$, but the case of $k=n$ is surprisingly tricky.
Attempted proof: We can write the rank-2 matrix $A_0$ as the following product: $$ A_0 = \pmatrix{ 1&1&\cdots & 1\\ 0&n&\cdots & n^2-n}^T \pmatrix{ 1&2&\cdots&n\\ 1&1&\cdots&1} $$ The modified matrix can be written as $$ A = {\underbrace{\pmatrix{ 1&1&\cdots & 1\\ 0&n&\cdots & n^2-n\\ 1&0&\cdots & 0\\ 0&1&\ddots\\ &&\ddots}}_B}^T \underbrace{ \pmatrix{ 1&2&\cdots&n\\ 1&1&\cdots&1\\ 1&-1&0&\cdots\\ &1&-1&\ddots\\ &&\ddots&\ddots} }_C $$ Where both matrices in the product ($B$ and $C$) $k$ rows. It suffices to show that $B$ and $C$ have rank $k$.
The matrix $B$ can be written in the block-matrix form $$ B = \pmatrix{B_{11} & B_{12}\\I_{k-2} & 0}. $$ We can apply block-row operations to get $$ B \leadsto \pmatrix{0 & B_{12}\\I_{k-2} & 0} \leadsto \pmatrix{I_{k-2} & 0\\ 0 & B_{12}}. $$ Thus, the rank of $B$ is equal to $(k-2) + \operatorname{rank}(B_{12})$, where $$ B_{12} = \pmatrix{ 1&1&\cdots & 1\\ (k-2)n & (k-1)n & \cdots & (n-1)n} $$ Now, we need only remark that $B_{12}$ has linearly independent rows and therefore has rank $2$.
Similarly, $C$ can be written as $$ C = \pmatrix{C_{11} & C_{12}\\ U & C_{22}}, $$ where $U$ is upper-triangular and invertible. We can apply block-row operations to get $$ \pmatrix{C_{11} & C_{12}\\ U & C_{22}} \leadsto \pmatrix{0 & C_{12} - C_{11}U^{-1}C_{22}\\ U & C_{22}}, $$ so it suffices to show that $C_{12} - C_{11}U^{-1}C_{22}$ has rank 2. We have $$ U^{-1} = \pmatrix{ 1 & 1 & \cdots & 1\\ 0 & 1 & \ddots & \vdots\\ \vdots & \ddots & \ddots & 1\\ 0 & \cdots & 0 & 1}, \quad C_{11} = \pmatrix{ 1&2&\cdots&k-2\\ 1&1&\cdots&1}, \\C_{12} = \pmatrix{ k-1 & k & \cdots & n\\ 1 & 1 & \cdots & 1 }, \quad C_{22} = \pmatrix{&&&\\ \\ -1}. $$ Notably, $C_{22}$ has all zero entries except for in its first column. Thus, the same is true for the product $C_{11}U^{-1}C_{22}$. Thus, all columns of $C_{12} - C_{11}U^{-1}C_{22}$ except the first are equal to that of $C_{12}$. Thus, to show that $C_{12} - C_{11}U^{-1}C_{22}$ has linearly independent rows, it suffices to show that the matrix attained by removing the first column of $C_{12} - C_{11}U^{-1}C_{22}$ has independent rows, which is to say that the matrix attained by removing the first column of $C_{12}$ has linearly independent columns. That is, it suffices to note that the matrix $$ \pmatrix{ k & \cdots & n\\ 1 & \cdots & 1} $$ has independent columns, which is indeed true for $k < n$.