Let us replace the value "two" from the statement by a general value $1\leq \ell \leq n$. Clearly the size of a convenient set $S$ is at least $\ell n$, and indeed there exist such convenient sets with $|S| = \ell n$. Let us consider a tripartite graph, having as classes of vertices the rows $R = \{r_1,r_2,\ldots,r_n\}$, the columns $C = \{c_1,c_2,\ldots,c_n\}$, and the elements of $M$. An edge $xr_i$ exists if $\left |r_i \cap (M \setminus \{x\})\right| < \ell$, and an edge $xc_j$ exists if $\left |c_j \cap (M \setminus \{x\})\right| < \ell$. Clearly the following relations hold: $1\leq \deg x \leq 2$, $\deg r_i\in \{0, \ell\}$, $\deg c_j \in \{0, \ell\}$, $\displaystyle \sum_{i=1}^n \deg r_i + \sum_{j=1}^n \deg c_j = \sum_{x\in M} \deg x$. It then follows $\displaystyle \sum_{x\in M} \deg x \geq \sum_{x\in M} 1 = |M|$. Let us now start with the simplest case, for $\ell = 1$, $n\geq 2$. A model $M$ with $2(n-1)$ elements is obtained by taking the first $n-1$ squares from row $r_1$ and the last $n-1$ squares from column $c_n$. To prove that no set $M$ can have more elements, assume $|M| \geq 2n-1$. Since $\displaystyle \sum_{i=1}^n \deg r_i + \sum_{j=1}^n \deg c_j = \sum_{x\in M} \deg x \geq |M| \geq 2n-1$, it follows that one of the two sums in the LHS, say that over the rows, is at least $\lceil (2n-1)/2\rceil = n$. Since on the other hand $\displaystyle \sum_{i=1}^n \deg r_i \leq \sum_{i=1}^n 1 = n$, it follows $\displaystyle \sum_{i=1}^n \deg r_i = n$, with all $\deg r_i = 1$. But that means $2n-1 \leq |M| = n$, absurd for $n\geq 2$. Thus $m=2(n-1)$. The case of the problem is $\ell = 2$, $n\geq 5$. A model $M$ with $4(n-2)$ elements is obtained by taking the first $n-2$ squares from rows $r_1, r_2$ and the last $n-2$ squares from columns $c_{n-1},c_n$. To prove that no set $M$ can have more elements, assume $|M| \geq 4n-7$. Since $\displaystyle \sum_{i=1}^n \deg r_i + \sum_{j=1}^n \deg c_j = \sum_{x\in M} \deg x \geq |M| \geq 4n-7$, it follows that one of the two sums in the LHS, say that over the rows, is at least $\lceil (4n-7)/2\rceil = 2n-3$. Consider the subset $R' \subset R$ of those rows containing exactly two elements from $S$. That means $\deg r = 2$ for $r \in R'$, and $\deg r = 0$ for $r\in R\setminus R'$, hence $\displaystyle \sum_{i=1}^n \deg r_i = \sum_{r \in R'} \deg r = \sum_{r \in R'} 2 = 2|R'| \geq 2n-3$, but then $\displaystyle \sum_{i=1}^n \deg r_i \geq 2n-2$, from considerents of parity, so $|R'| \geq n-1$. But we cannot have $|R'| = n$, as that means $R'= R$, and then $4n-7 \leq M = 2n$, absurd for $n\geq 5$, so $|R'| = n-1$. The only row from $R\setminus R'$ can contain at most $n-1$ elements from $M$ (if $n$, that means that, counting by columns, $|M| = 2n$, absurd), therefore $4n-7 \leq M \leq 2(n-1) + (n-1) = 3n-3$, absurd for $n\geq 5$. This is where it is seen the reason behind the condition $n\geq 5$, since for $n=4$ the answer is given by a different model, with $9 > 4(n-2) = 8$ elements (while for $n=3$ the model has $6 > 4(n-2) = 4$ elements, and for $n=2$ the model has $4 > 4(n-2) = 0$ elements). Thus $\boxed{m=4(n-2)}$. For larger values of $\ell$, similar arguments (involving some extra complications) lead to $m = 2\ell(n-\ell)$, for $n \geq 3\ell - 1$.