Bariz - Problem

Let $ a_1, a_2,\ldots ,a_8$ be $8$ distinct points on the circumference of a circle such that no three chords, each joining a pair of the points, are concurrent. Every $4$ of the $8$ points form a quadrilateral which is called a quad. If two chords, each joining a pair of the $8$ points, intersect, the point of intersection is called a bullet. Suppose some of the bullets are coloured red. For each pair $(i j)$, with $ 1 \le i < j \le 8$, let $r(i,j)$ be the number of quads, each containing $ a_i, a_j$ as vertices, whose diagonals intersect at a red bullet. Determine the smallest positive integer $n$ such that it is possible to colour $n$ of the bullets red so that $r(i,j)$ is a constant for all pairs $(i,j)$.

Note that coloring $1$ bullet red, say $A_j A_k \cap A_xA_y$, increments each of the $\binom42$ $r(s, t)$'s with $s, t \in \{j, k, x, y\}$ by one. Therefore, the problem is equivalent to finding the minimal $n$ such that we can pick $n$ quadruples of $A_i$'s such that every edge $A_iA_j$ is contained in the same number $z$ of quadruples. Observe that by an easy double-counting argument, we have that $\binom82 \cdot z = \binom42 \cdot n$, and so we clearly obtain that $3 | z$ by mod $3$. This means that $n \ge \frac{\binom82 \cdot 3}{\binom42} = 14.$ All that remains is to construct an example with $14$ quadruples. To do so, simply use $\{1, 2, 3, 4\}, \{1, 2, 5, 6\}, \{1, 2, 7, 8\}, \{3,4,5,6\}, \{3,4,7,8\}, \{5,6,7,8\},$ and all of the $\frac{2^4}{2} = 8$ quadruples in $\{1, 2\} \times \{3, 4\} \times \{5, 6\} \times \{7, 8\}$ which have the property that the sum of all four elements is even.