$m = \lfloor \frac{n+3}{2} \rfloor$.

Put these $n^2$ numbers $x_{i,j}$ in an $n \times n$ table: call $x_{i,j}$ a "row pivot" if $x_{i,j} = \max\{x_{i,1}, x_{i,2}, \dots, x_{i,j}\}$; call $x_{i,j}$ a "column pivot", if $x_{i,j} = \max\{x_{1,j}, x_{2,j}, \dots, x_{i,j}\}$.

First, prove $m \ge \lfloor \frac{n+3}{2} \rfloor$. Consider all $(n-1)^2$ numbers $x_{i,j}$ ($2 \le i, j \le n$): every column (from $2$ to $n$) contains at most $m-1$ column pivots (by condition (3), and notice that $x_{1,j}$ ($2 \le j \le n$) is a pivot); every row (from $2$ to $n$) contains at most $m-1$ row pivots (by condition (2), and notice that $x_{i,1}$ ($2 \le i \le n$) is a pivot). Due to condition (1), every number is a pivot (row or column or both), and hence
$$(n-1)(m-1) + (n-1)(m-1) \ge (n-1)^2. \quad \textcircled{1}$$
We claim the equality in (1) cannot be attained. Let $x_{i_0,j_0} = \max x_{i,j}$ ($2 \le i, j \le n$). If $x_{i_0,j_0}$ is both row and column pivots, then $x_{i_0,j_0}$ is counted twice on the left-hand side of (1), and the equality cannot hold. Suppose $x_{i_0,j_0}$ is not a row pivot. Then $x_{i_0,1} > x_{i_0,j_0}$ and there are no other row pivots in this row, the equality in (1) cannot hold. Similarly, if $x_{i_0,j_0}$ is not a column pivot, then there are no other column pivots in this column, and the equality in (1) cannot hold. Therefore, $2(m-1) > n-1$, $m \ge \lfloor \frac{n+3}{2} \rfloor$.

Next, construct tables for which $m = \lfloor \frac{n+3}{2} \rfloor$. When $n = 2t$, divide the $n \times n$ table into four $t \times t$ tables as follows:

Take $\varepsilon_1, \varepsilon_2 > 0$ sufficiently small, such that $\varepsilon_1 < \frac{1}{2n}\varepsilon_2 < \frac{1}{4n^2}$. For $k=1, 4$, let the number in row $i$ and column $j$ of $\Omega_k$ be $k - i\varepsilon_1 + j\varepsilon_2$. Note that in $\Omega_1$ and $\Omega_4$, the numbers decrease from the top right to the bottom left. For $k = 2, 3$, let the number in row $i$ and column $j$ of $\Omega_k$ be $k + i\varepsilon_1 - j\varepsilon_2$. Note that in $\Omega_2$ and $\Omega_3$ the numbers decrease from the bottom left to the top right.

It is easy to check that all numbers in $\Omega_1$ and $\Omega_4$ are row pivots, and all numbers in $\Omega_2$ and $\Omega_3$ are column pivots. In addition, except for the first and the $t+1$ rows, there are no other column pivots in $\Omega_1$ and $\Omega_4$; except for the first and the $t+1$ columns, there are no other row pivots in $\Omega_2$ and $\Omega_3$. Hence, conditions (2) and (3) are fulfilled, and $m = t+1$.

When $n = 2t - 1$, take the construction for $n = 2t$ as above and remove the rightmost column and the bottom row. $\square$