Example 4.3.3 $n$ points, no three of which are collinear, are connected by line segments, and these segments are colored with two colors, red and blue. If for any coloring, there must exist 12 monochromatic triangles, find the minimum value of $n$.


Solution: $n$ points can form $\mathrm{C}_{n}^{3}$ triangles. Excluding at least 12 monochromatic triangles, there are at most $\mathrm{C}_{n}^{3}-12$ bichromatic triangles, hence there are at most $2\left(\mathrm{C}_{n}^{3}-12\right)$ bichromatic angles. From any point $A_{i}$, $n-1$ line segments can be drawn, with $r_{i}$ red lines and $n-1-r_{i}$ blue lines. The number of bichromatic angles with $A_{i}$ as a vertex is $r_{i}\left(n-1-r_{i}\right) \leqslant\left(\frac{n-1}{2}\right)^{2}$. The total number of bichromatic angles is $\sum_{i=1}^{n} r_{i}\left(n-1-r_{i}\right) \leqslant n\left(\frac{n-1}{2}\right)^{2}$. If $n$ is odd, there is a coloring method that makes the equality hold. That is, there can be $n\left(\frac{n-1}{2}\right)^{2}$ bichromatic angles. Therefore, it must be that $n\left(\frac{n-1}{2}\right)^{2} \leqslant 2\left(\mathrm{C}_{n}^{3}-12\right)$, solving this gives $n \geqslant 9$.
When $n=8$, through each point 7 lines are all colored with three red and four blue, the number of bichromatic triangles is $\frac{1}{2} \sum_{i=1}^{8} r_{i}\left(7-r_{i}\right) = \frac{1}{2} \times 8 \times (3 \times 4) = 48$, $\mathrm{C}_{8}^{3}-48=8$, there are only 8 monochromatic triangles. Therefore, the answer to this problem is $n=9$.