63rd Czech and Slovak Mathematical Olympiad

For any $k\in \{1,2,...,n\}$ denote by $p_k$ the number of visitors in the $k$-th row. The stated condition on given $i$ and $j$ implies that the number of friendly pairs $(A,B)$, where $A$ and $B$ are from the $i$-th row and from $j$-th row respectively, is equal to the product $j{p}_i$. Interchanging the indices $i$ and $j$, we conclude that the same number of friendly pairs $(A,B)$ equals $i{p}_j$. Thus $j{p}_i=i{p}_j$ or $p_i:p_j=i:j$, and therefore, all the numbers $p_k$ must be proportional as follows: $$p_1:p_2:\cdots :p_n=1:2:\cdots :n.$$ Let us show that under this proportionality the visitors can be friendly in such a way which ensures the property under consideration. Thus assume that for some positive integer $d$, the equality $p_k=kd$ holds with any $k\in \{1,2,...,n\}$. Let us

start with the case $d=1$ when the numbers of visitors in single rows are successively $1,2,...,n$. Then the stated property holds true if (and only if) any two visitors — taken from distinct rows in the whole auditorium — are friends. In the case when $d>1$, let us divide all the visitors into $d$ groups $G_1,G_2,...,G_d$ so that for arbitrary $k=1,2,...,d$, the numbers of visitors from the group $G_k$ in single rows are successively $1,2,...,n$. It is evident that the stated property holds true under the following condition: two visitors are friends if and only if they belong to the same group $G_k$. It follows from the preceding that our task is to find such integer values of $n$, $n\ge 4$, for which there exists a positive integer $d$ satisfying the equation $$d+2d+\cdots +nd=234\text{or}dn(n+1)=468.$$ Thus we look for all divisors $468=2^2\cdot 3^2\cdot 13$ which are of the form $n(n+1)$. Inequality $22\cdot 23>468$ implies that $n<22$ and hence $n\in \{4,6,9,12,13,18\}$. It is easy to see that the only satisfactory $n$ equals $12$ (which corresponds to $d=3$).

Answer. The unique solution is $n=12$.