1.3. In how many ways can $n^{2}$ distinct real numbers be arranged into an $n \times n$ array $\left(a_{i j}\right)$ such that $\max _{j} \min _{i} a_{i j}=\min _{i} \max _{j} a_{i j}$ ?
1.3 If $\max _{j} \min _{i} a_{i j}=\min _{i} \max _{j} a_{i j}=a_{\alpha \beta}$, then clearly $a_{\alpha \beta}$ is at once the largest number in the $\alpha$ th row and the smallest numbers in the $\beta$ th column, and hence
$$
a_{\alpha j}a_{\alpha \beta}$ for all $i \neq \alpha$ would imply that $\max _{j} \min _{i} a_{i j}=a_{\alpha \beta}=\min _{i} \max _{j} a_{i j}$. To obtain a required configuration, it is therefore necessary and sufficient to choose any $2 n-1$ of the given $n^{2}$ numbers, say $x_{1}<x_{2}<\cdots<x_{2 n-1}$. Put $x_{n}$ anywhere in the array. Then put $x_{1}, x_{2}, \ldots, x_{n-1}$ in the same row as $x_{n}$ and put $x_{n+1}, x_{n+2}, \ldots, x_{2 n-1}$ in the same column as $x_{n}$. The remaining $n^{2}-2 n+1$ numbers can be used to fill up the remaining $n^{2}-2 n+1$ positions. Therefore, the total numbers of such configurations is
$$
\binom{n^{2}}{2 n-1} \cdot n^{2} \cdot[(n-1)!]^{2} \cdot\left(n^{2}-2 n+1\right)!=\frac{\left(n^{2}\right)!(n!)^{2}}{(2 n-1)!}
$$