Saudi Arabian Mathematical Competitions, 2012

The answer is $n! \cdot (n-1)^n$. This is achievable by choosing the numbers $n-1, 2(n-1), 3(n-1), \dots, n(n-1)$, which are at positions $(n, 1), (n-1, 2), (n-2, 3), \dots, (1, n)$. To show this is the best possible, we begin with a lemma.

Lemma. If $a < b < c < d$ are real numbers such that $a+d = b+c$, then $bc > ad$.

Proof. We have $(b-a)(c-a) > 0$, which implies
$$bc + (a - b - c)a > 0.$$
Since $a - b - c = -d$, this simplifies to $bc > ad$ as desired. $\square$

Let $f(i, j) = (i - 1) + n(j - 1)$ be the number in row $i$ and column $j$. Suppose we have chosen the $n$ numbers $f(1, j_1), f(2, j_2), \dots, f(n, j_n)$. If the $j_i$ sequence is strictly decreasing, then $j_i = n + 1 - i$ for all $i$ and we get a configuration in the first paragraph. Else there exists $k, m$ such that $k < m$ and $j_k < j_m$. Notice that
$$f(k, j_k) < f(m, j_k) < f(k, j_m) < f(m, j_m).$$
Moreover,
$$\begin{aligned} f(k, j_k) + f(m, j_m) &= f(m, j_k) + f(k, j_m) \\ &= (k + m - 2) + n(j_k + j_m - 2). \end{aligned}$$
So by the lemma, we have that
$$f(k, j_k)f(m, j_m) < f(k, j_m)f(m, j_k).$$
Therefore we increase the product of the $n$ numbers by changing $f(k, j_k)$ and $f(m, j_m)$ to $f(k, j_m)$ and $f(m, j_k)$, so our original arrangement could not have been a maximum.