China Southeastern Mathematical Olympiad

First, we consider the maximal number of triangles with no common side. Consider the maximal number of triangles with no common side pairwise. There are $C_8^2=28$ chords by connecting eight points. If each chord only belongs to one triangle, then these chords can only form $r\le \lfloor \frac{28}{3}\rfloor =9$ triangles with no common side pairwise. But if there are nine such triangles, then there are 27 vertices. So, there is one point in eight points, which is the common vertex of four triangles.



Suppose that point is $A_8$, then eight edges are connected to seven points $A_1,A_2,\dots ,A_7$. So, there must exist an edge $A_8{A}_k$, which is the common side of two triangles, which is a contradiction. So $r\le 8$.

On the other hand, when $r=8$, we can make such eight triangles, see figure. Denote the triangles by three vertices as: $(1,2,8)$, $(1,3,6)$, $(1,4,7)$, $(2,3,4)$, $(2,5,7)$, $(3,5,8)$, $(4,5,6)$ and $(6,7,8)$. So the minimal number of $n$ is 9.