One of the inductive approaches at RMM 2017/5 (e.g. post 11) is directly mimicked here. Both problems become of pretty much the same difficulty if one says in advance in the RMM problem that the answer is the same regardless of the locations of the blocks.

Answer is $2n-2$

https://pregatirematematicaolimpiadejuniori.wordpress.com/shl-2023/ OFFICIAL SOLUTION

Enjoyed while solving this. Claim: $\text{the number of rooks} \le 2n-2$ Proof: It's known that we have $2$ blocks these are on {top left,top right,bottom left, bottom right} let's call them expectional blocks and call the row expectional that contains expectional block . So we can place $2$ rooks on each row that isn't expectional row. We have $n-2$ non-expectional row and we have $2$ rooks at max on each of them. Also, we can place $1$ rook on each expectional row. That means we have $2n-2$ rooks at maxiumum. Let's prove desired number is $2n-2$. For $2 \cdot 2$ and $3 \cdot 3$ we can easily deduce the answer is $2n-2$. So let's perform relation between $n$ and $n-2$. Assume our answer is $2n-2$ for $n \cdot n$. We can create $(n+2) \cdot (n+2)$ grid by adding $n+2$ cells to every side of $n \cdot n$ grid. Define coordinates of grid like $a_0,a_2,....,a_{n+1}$ as $x$ axis and $b_0,b_2,....,b_{n+1}$ as $y$ axis. If our expectional blocks coordinated as $(a_1,b_1)$ and $(a_n,b_n)$ we can add $4$ rocks $(a_0,b_1),(a_1,b_0),(a_n,b_n+1),(a_n+1,b_n)$ which would give us $2n+2$ for $(n+2) \cdot (n+2)$ grid. For expectional blocks that are coordinated as $(a_n,b_1),(a_1,b_n)$ we can do same thing which would gives us $2n+2$. $\blacksquare$

The answer is $2n-2$ for $n$ and the configuration for $n$ should look like something like this: $\begin{bmatrix} R &\boxed{} & \\ \boxed{}& R& \\ R & \boxed{}& R \end{bmatrix}$ The cases $n=1,2$ are trivial we shall prove that if we add a row and a column we can always place 2 rooks on the board.When we add the column and the row:: $\begin{bmatrix} R &\boxed{} & \\ \boxed{}& R& \\ R & \boxed{}& R \end{bmatrix}\to \begin{bmatrix} R& & & \\ \boxed{}& R& & \\ R& \boxed{}& R& \\ \boxed{}& R& \boxed{} & R \end{bmatrix}$. We place the first rook at $Int(R_k:C_k)$ and move two blocks and add the other rook and place a block between these roots so $N_k(R)+2=N_{k+1}(R)$ hence proven.