Question 3 Given $n$ points $p_{1}, p_{2}, \cdots, p_{n}$ in the plane, with no three points collinear. Each point $p_{i}(i=1,2, \cdots, n)$ is arbitrarily colored red or blue. Let $S$ be a set of some triangles with vertex set $\left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, and has the property: for any two line segments $p_{i} p_{j}$ and $p_{h} p_{k}$ in the graph, the number of triangles in $S$ that have $p_{i} p_{j}$ as a side is equal to the number of triangles in $S$ that have $p_{h} p_{k}$ as a side. Find the smallest positive integer $n$, such that in the set $S$ there are always two triangles, each with three vertices of the same color.
(2007, China National Training Team Test)
When $n \in\{1,2, \cdots, 5\}$, it is clearly impossible to satisfy the conditions of the problem.

When $n=6$, color $p_{1}, p_{2}, p_{3}$ red, and $p_{4}, p_{5}, p_{6}$ blue, and consider the triangles:
$\triangle p_{1} p_{2} p_{3}, \triangle p_{1} p_{3} p_{4}, \triangle p_{1} p_{4} p_{5}, \triangle p_{1} p_{5} p_{6},$
$\triangle p_{1} p_{2} p_{6}, \triangle p_{2} p_{3} p_{5}, \triangle p_{2} p_{4} p_{5}, \triangle p_{2} p_{4} p_{6},$
$\triangle p_{3} p_{4} p_{6}, \triangle p_{3} p_{5} p_{6}$.
It is easy to verify that for any line segment $p_{i} p_{j}$ $(1 \leqslant i<j \leqslant 6)$, there are exactly two triangles, but only $\triangle p_{1} p_{2} p_{3}$ has all three vertices of the same color.

Therefore, the set $S$ of these ten triangles does not satisfy the conditions of the problem.

When $n=7$, color $p_{1}, p_{2}, p_{4}$ red, and $p_{3}, p_{5}, p_{6}, p_{7}$ blue, and consider the triangles formed by the cyclic permutation arrangement:
$\triangle p_{1} p_{2} p_{4}, \triangle p_{2} p_{3} p_{5}, \triangle p_{3} p_{4} p_{6}, \triangle p_{4} p_{5} p_{7},$
$\triangle p_{5} p_{6} p_{1}, \triangle p_{6} p_{7} p_{2}, \triangle p_{7} p_{1} p_{3}$.
It is easy to verify that the 21 edges of the seven triangles are exactly the 21 edges of the complete graph of order 7. But among the seven triangles, only $\triangle p_{1} p_{2} p_{4}$ has all three vertices of the same color.

Therefore, the set $S$ of these seven triangles does not satisfy the conditions of the problem.

When $n=8$, call the line segment connecting two points of the same color a same-color edge, and the line segment connecting two points of different colors a different-color edge. Thus, there are at most 16 different-color edges in the graph. At this point, there are four red and four blue points among the eight vertices, with 12 same-color edges, and the graph has 28 edges in total.

Since 3 does not divide $28, 56, 112, 140$, but divides $84, 168$, to satisfy the conditions of the problem, each edge can only be a common edge of three triangles or six triangles.

If $k=3$, i.e., each edge is a common edge of three triangles, then $|S|=28$.

Note that each different-color triangle has exactly two different-color edges, and there are at most 48 different-color edges in the graph (each edge is counted three times). Therefore, there are at most 24 different-color triangles. Thus, the set $S$ must contain at least four same-color triangles.

If $k=6$, then $|S|=56$. Similarly, there must be at least eight same-color triangles.
This indicates that when $n=8$, the conditions of the problem are satisfied.
In summary, the smallest positive integer $n=8$.