Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Suppose, if possible, that the positive divisors of some integer $n > 1$ may be arranged as required in an $k \times \ell$ rectangular array, where $k$ is the number of rows, and $\ell$ is the number of columns; clearly, $k$ and $\ell$ must both be greater than $1$.
Let $s$ be the common value of the row sums, and notice that $s \ge n + 1$.
Let $d_i$ be the largest divisor on the $i$-th row, $i = 1, \dots, k$. Assume, without any loss, $d_1 > \dots > d_k$, to infer that the $n/d_i$, $i = 1, \dots, k$, form a strictly increasing $k$-element string of positive integers, so $n/d_k \ge k$; that is, $d_k \le n/k$.
Since $d_k$ is maximal along the $k$-th row,
$$d_k \ge s/\ell > n/\ell,$$
so $\ell > k$, by the preceding (in fact, $d_k > s/\ell$, since the divisors are pairwise distinct).
Mutatis mutandis, $k > \ell$, and we reach a contradiction.