THE 68th 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, $d_1>\cdots >d_k$, 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.