A set of (unit) squares of a $n\times n$ table is called [i]convenient[/i] if each row and each column of the table contains at least two squares belonging to the set. For each $n\geq 5$ determine the maximum $m$ for which there exists a [i]convenient [/i] set made of $m$ squares, which becomes in[i]convenient [/i] when any of its squares is removed.
1. **Generalization and Initial Setup:**
   - We generalize the problem by replacing the value "two" with a general value \(1 \leq \ell \leq n\).
   - A convenient set \(S\) must have at least \(\ell n\) elements, and such sets exist with \(|S| = \ell n\).

2. **Graph Representation:**
   - Consider a tripartite graph with vertex classes: rows \(R = \{r_1, r_2, \ldots, r_n\}\), columns \(C = \{c_1, c_2, \ldots, c_n\}\), and 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 \(\left|c_j \cap (M \setminus \{x\})\right| < \ell\).

3. **Degree Constraints:**
   - Clearly, \(1 \leq \deg x \leq 2\), \(\deg r_i \in \{0, \ell\}\), \(\deg c_j \in \{0, \ell\}\).
   - The sum of degrees: \(\sum_{i=1}^n \deg r_i + \sum_{j=1}^n \deg c_j = \sum_{x \in M} \deg x\).
   - It follows that \(\sum_{x \in M} \deg x \geq |M|\).

4. **Case \(\ell = 1\):**
   - For \(n \geq 2\), 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 no set \(M\) can have more elements, assume \(|M| \geq 2n-1\).
   - Since \(\sum_{i=1}^n \deg r_i + \sum_{j=1}^n \deg c_j = \sum_{x \in M} \deg x \geq |M| \geq 2n-1\), one of the sums on the LHS, say over the rows, is at least \(\lceil (2n-1)/2 \rceil = n\).
   - Since \(\sum_{i=1}^n \deg r_i \leq \sum_{i=1}^n 1 = n\), it follows \(\sum_{i=1}^n \deg r_i = n\), with all \(\deg r_i = 1\).
   - This implies \(2n-1 \leq |M| = n\), which is absurd for \(n \geq 2\).
   - Thus, \(m = 2(n-1)\).

5. **Case \(\ell = 2\):**
   - For \(n \geq 5\), 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 no set \(M\) can have more elements, assume \(|M| \geq 4n-7\).
   - Since \(\sum_{i=1}^n \deg r_i + \sum_{j=1}^n \deg c_j = \sum_{x \in M} \deg x \geq |M| \geq 4n-7\), one of the sums on the LHS, say 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 \(S\). This means \(\deg r = 2\) for \(r \in R'\), and \(\deg r = 0\) for \(r \in R \setminus R'\).
   - Hence, \(\sum_{i=1}^n \deg r_i = \sum_{r \in R'} \deg r = \sum_{r \in R'} 2 = 2|R'| \geq 2n-3\).
   - From considerations of parity, \(\sum_{i=1}^n \deg r_i \geq 2n-2\), so \(|R'| \geq n-1\).
   - We cannot have \(|R'| = n\), as that means \(R' = R\), and then \(4n-7 \leq |M| = 2n\), which is absurd for \(n \geq 5\).
   - Thus, \(|R'| = n-1\). The only row from \(R \setminus R'\) can contain at most \(n-1\) elements from \(M\) (if \(n\), that means, counting by columns, \(|M| = 2n\), absurd).
   - Therefore, \(4n-7 \leq |M| \leq 2(n-1) + (n-1) = 3n-3\), which is absurd for \(n \geq 5\).

6. **Conclusion for \(\ell = 2\):**
   - Thus, \(\boxed{m = 4(n-2)}\).

7. **General Case for Larger \(\ell\):**
   - For larger values of \(\ell\), similar arguments (involving some extra complications) lead to \(m = 2\ell(n-\ell)\), for \(n \geq 3\ell - 1\).