8. Given 8 points $A_{1}, A_{2}, \cdots, A_{8}$ on a circle. Find the smallest positive integer $n$, such that among any $n$ triangles with these 8 points as vertices, there must be two triangles that share a common side.
(Tao Pingsheng, problem contributor)
8. First, consider the maximum number of triangles with no common edges.

Connecting every pair of eight points yields $\mathrm{C}_{8}^{2}=28$ chords.

If each chord belongs to only one triangle, then these chords can form at most $r \leqslant\left[\frac{28}{3}\right]=9$ triangles with no common edges. However, if there are nine such triangles, they would have a total of 27 vertices, meaning that one of the eight points (let's say $A_{8}$) must belong to at least four triangles. The opposite sides of the four triangles sharing vertex $A_{8}$ are all lines connecting two points among $A_{1}, A_{2}, \cdots, A_{7}$. Therefore, one point (let's say $A_{k}$) must appear twice, which means the corresponding two triangles will share a common edge $A_{8} A_{k}$, leading to a contradiction. Therefore, $r \leqslant 8$.

On the other hand, it is possible to construct eight triangles such that any two of them have no common edges.

Notice that such eight triangles produce a total of 24 vertices, and each point belongs to at most three triangles. Hence, each point belongs to exactly three triangles, meaning each point has a degree of 6.

For simplicity, as shown in Figure 3, take the eight equally spaced points on the circumference as the eight vertices of graph $G$, construct an 8-vertex complete graph, and then remove four diameters. Thus, we get 24 edges, and each point belongs to six edges. Among the triangles formed by these edges, select eight isosceles triangles:
$$
(1,2,3),(3,4,5),(5,6,7),(7,8,1)
$$

and $(1,4,6),(3,6,8),(5,8,2),(7,2,4)$, which have no common edges (each set of four triangles can be obtained by rotating one of the triangles around the center appropriately).
Therefore, $r=8$.
Thus, the required minimum value $n=r+1=9$.