Example 2: There are 18 points on a plane, no three of which are collinear. Each pair of points is connected by a line segment, which is colored either red or blue, with each line segment being only one color. It is known that a certain point $A$ has an odd number of red line segments emanating from it, and the number of red line segments emanating from the other 17 points are all different.
(1) Find the number of red triangles in this graph;
(2) Find the number of triangles in this graph that have exactly two red sides. (36th IMO Shortlist)
Remove all blue edges from the graph, obtaining a simple graph $G$. By the given conditions, $d(A)$ is odd, and the degrees of other non-$A$ points belong to $\{0,1,2, \cdots, 17\}$. It is easy to see that points with degree 0 and 17 cannot coexist. Therefore, the degrees of the points are either $0,1,2, \cdots, 16$ or $1,2, \cdots, 17$.

Notice that $\sum d(x)=d(A)+0+1+2+\cdots+16$ is odd, which is a contradiction. Therefore, the degrees of the vertices can only be $1,2, \cdots, 17$. Let the vertices be denoted as $V_{1}, V_{2}, \cdots, V_{17}$, where $d\left(V_{i}\right)=i$ $(i=1,2, \cdots, 17)$.

Since $d\left(V_{17}\right)=17$, $V_{17}$ is connected to $V_{1}$. Also, since $d\left(V_{1}\right)=1$, $V_{1}$ is only connected to $V_{17}$, so $V_{16}$ is not connected to $V_{1}$. However, $d\left(V_{16}\right)=16$, so $V_{16}$ is connected to $V_{2}, V_{3}, \cdots, V_{17}$. Continuing this pattern, $V_{i}$ is connected to $V_{18-i}, V_{19-i}, \cdots, V_{17}$ $(i=1,2, \cdots, 8)$, and $V_{18-i}$ is connected to all points except $V_{1}, V_{2}, \cdots, V_{i-1}$ $(i=1,2, \cdots, 8)$. Now consider $V_{9}$, which should be connected to $A, V_{17}, V_{16}, V_{15}, \cdots, V_{10}$. Since $V_{9}$ is connected to $A$, $V_{1}, V_{2}, \cdots, V_{8}$ are not connected to $A$, and $V_{9}, V_{10}, \cdots, V_{17}$ are connected to $A$, so $d(A)=9$.

Let $M=\left\{V_{1}, V_{2}, \cdots, V_{8}\right\}$ and $N=\left\{A, V_{9}, V_{10}, \cdots, V_{17}\right\}$. Then the points in $M$ are not connected to each other, while the points in $N$ are fully connected, and each point $V_{i}$ in $M$ is connected to $i$ points in $N$.
(1) The problem is equivalent to finding the number of triangles in graph $G$. First, there are $\mathrm{C}_{10}^{3}$ triangles in $N$. Additionally, for any point $V_{i}$ in $M$, it is connected to $i$ points in $N$, so it can form $\mathrm{C}_{i}^{2}$ triangles. The number of such triangles is $\sum_{i=1}^{8} \mathrm{C}_{i}^{2}=\mathrm{C}_{9}^{3}$. Therefore, the total number of red triangles is $\mathrm{C}_{10}^{3}+\mathrm{C}_{9}^{3}=204$.
(2) Calculate the total number of red angles $S$ in two ways.

On one hand, calculate $S$ in $G$. Since point $V_{i}$ has $i$ edges $(i=2,3, \cdots, 17)$, the number of red angles with $V_{i}$ as the vertex is $\mathrm{C}_{i}^{2}$. Therefore, $S=\mathrm{C}_{9}^{2}$ (red angles from $A$) $+\sum_{i=2}^{17} \mathrm{C}_{i}^{2}=\mathrm{C}_{9}^{2}+\mathrm{C}_{18}^{3}$.

On the other hand, each red angle is in some triangle. Suppose there are $x$ triangles with exactly two red edges, each such triangle has exactly one red angle, contributing $x$ red angles. From (1), there are 204 red triangles, each with 3 red angles, contributing $204 \times 3=612$ red angles. Additionally, other triangles have no red angles. Therefore, $S=x+612$. By $\mathrm{C}_{9}^{2}+\mathrm{C}_{18}^{3}=S=x+612$, we solve for $x=240$.