Example 11 Find the smallest positive integer $n$, such that when $K_{n}$ is colored in two colors in any way, there always exist two monochromatic triangles, each sharing exactly one common edge.
(1993 China National Training Team Practice Problem)
In Fig. 7-12(a) and (b), for clarity, we draw the blue and red lines in two separate figures. It is easy to see that there are 6 blue and 6 red, a total of 12 monochromatic triangles in the figure, with 36 edges in the figure, and each edge is one side of a monochromatic triangle. No two monochromatic triangles share a common edge, so the smallest positive integer \( n \geqslant 10 \).

On the other hand, in a 2-colored \( K_{10} \), assume there are \( p \) monochromatic triangles and \( q \) non-monochromatic triangles, then \( p+q=C_{10}^{3} \). Let \( x_{i} \) be the number of red edges starting from the \( i \)-th vertex \( p_{i} \), and \( 9-x_{i} \) be the number of blue edges \((i=1,2, \cdots, 10)\). Thus, the graph has \( \sum_{i=1}^{10} x_{i}\left(9-x_{i}\right) \) bichromatic angles. Since there are no bichromatic angles in monochromatic triangles, and each non-monochromatic triangle has exactly 2 bichromatic angles, we have \( 2 q=\sum_{i=1}^{10} x_{i}\left(9-x_{i}\right) \). Because \( x_{i}\left(9-x_{i}\right) \leqslant\left(\frac{x_{i}+9-x_{i}}{2}\right)^{2}=\frac{81}{4}=20 \frac{1}{4} \), and \( x_{i}\left(9-x_{i}\right) \) is an integer, so \( x_{i}\left(9-x_{i}\right) \leqslant 20 \), hence \( q=\frac{1}{2} \sum_{i=1}^{10} x_{i}\left(9-x_{i}\right) \leqslant \frac{1}{2} \times 10 \times 20=100 \), thus \( p=C_{10}^{3}-q \geqslant 120-100=20 \). That is, there are at least 20 monochromatic triangles in the graph, each triangle has 3 edges, so there are at least 60 edges in total, but \( K_{10} \) has only \( C_{10}^{2}=45 \) line segments, so there must be two monochromatic triangles that share exactly one common edge.
In summary, the smallest positive integer \( n=10 \).