Vietnam Mathematical Olympiad

Suppose that there are $n$ crosses which can be coloured so that the numbers $s(i,k)$ are all equal. Denote by $s$ the common value of these numbers. Since for each couple of vertices $A_i,A_k$ of the octagon, there exist exactly $s$ subquadrilaterals admitting $A_i,A_k$ as vertices and admitting a coloured cross as the point of intersection of diagonals, hence the number of coloured crosses (with iterations) is $s{C}_8^2$. Since for each coloured cross, there exist exactly $C_4^2$ couples of vertices of the octagon which define a same subquadrilateral admitting this coloured cross as the point of intersection of the diagonals, each coloured cross is counted $C_4^2$ times in the above mentioned numeration. Therefore, $$n=(s\cdot C_8^2):C_4^2=\frac{14s}{3}.$$ As $n$ is a positive integer, $s$ is divisible by $3$. So $s\ge 3$ and it implies $n\ge 14$. Now we describe a colouring of $14$ crosses so that numbers $s(i,k)$ are all equal. For this purpose, we shall denote a cross by a subquadrilateral admitting it as the point of intersection of its diagonals and we shall denote the subquadrilateral $A_i{A}_k{A}_p{A}_q$ by $(ikpq)$. We colour the following crosses: (1234), (1256), (1278), (1357), (1368), (1458), (1467); (2358), (2367), (2457), (2468), (3456), (3478), (5678). By direct verification for each couple of vertices of the octagon, we see that $s(i,k)=3$ for all $i,k\in \{1,2,3,\dots ,8\},i\ne k$. So the required least positive integer $n$ is $n_{\min }=14$.