SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Suppose that there are $m$ students in total. We number the students standing in the circle by $1,2,3,\dots ,m$, clockwise (starting with an arbitrary student). Denote by $a_1<a_2<a_3<\cdots <a_n$ the positions of the $n$ head students obtained in this numbering.

We consider an $n\times m$ table and fill numbers in its cells in the following way: For each $1\le i\le n$, in the $i^{\text{th}}$ row, we start with $a_i$ (filling in the first cell of this row with $a_i$) and continue with $$a_i+1,a_i+2,\dots ,m,1,2,\dots ,a_i-1$$ to the right. We will focus on the hat colors of the students in the first column.

$a_1$	$a_1+1$	$a_1+2$	$\dots$	$a_1-1$
$a_2$	$a_2+1$	$a_2+2$	$\dots$	$a_2-1$
$a_3$	$a_3+1$	$a_3+2$	$\dots$	$a_3-1$
$a_n$	$a_n+1$	$a_n+2$	$\dots$	$a_n-1$

At the first minute, each student in the first column wears his own hat.

At the second minute, based on the game rule, each student in the first column will wear the hat of the student in the same row but in the second column.

Similarly, at the $k^{\text{th}}$ minute, each student in the first column will wear the hat of the student in the same row but in the $k^{\text{th}}$ column, for each $1\le k\le m$.

Suppose, on the contrary, that we could never find 2 head students (in the first column) wearing hats of the same color. This would imply that at the beginning all the students in each column also wear hats of pairwise different colors, so they belong to distinct classes. Hence, the numbers of students of $n$ classes are the same, which is a contradiction to the hypothesis.

Therefore, at some moment, there are 2 head students wearing hats of the same color.