Example 5 Given a set of points $P=\left\{P_{1}, P_{2}, \cdots, P_{1924}\right\}$ on a plane, and no three points in $P$ are collinear. Divide all the points in $P$ into 83 groups arbitrarily, such that each group has at least three points, and each point belongs to exactly one group. Then, connect any two points in the same group with a line segment, and do not connect points in different groups, thus obtaining a pattern $G$. Different grouping methods result in different patterns. Let the number of triangles in pattern $G$ with vertices from points in $P$ be denoted as $m(G)$.
(1) Find the minimum value $m_{0}$ of $m(G)$;
(2) Let $G^{*}$ be a pattern such that $m\left(G^{*}\right)=m_{0}$. If the line segments (referring to segments with endpoints from points in $P$) in $G^{*}$ are colored with 4 colors, and each segment is colored exactly one color. Prove: There exists a coloring scheme such that after coloring $G^{*}$, there is no triangle with vertices from points in $P$ whose three sides are of the same color.
(1994, National High School Mathematics Competition)
Solution: (1) Let $m(G)=m_{0}, G$ is obtained from the groups $X_{1}, X_{2}, \cdots, X_{88}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots, 83$.
Let $\left|X_{i}\right|=x_{i}, i=1,2, \cdots, 83$, then we have
$$
x_{1}+x_{2}+\cdots+x_{83}=1994 \text {, }
$$

and $m_{0}=C_{x_{1}}^{3}+C_{x_{2}}^{3}+\cdots+C_{x_{83}}^{3}$.
We now prove that, for $1 \leqslant i \neq j \leqslant 83$, we have
$$
\left|x_{i}-x_{j}\right| \leqslant 1 \text {. }
$$

In fact, if there exist $i, j (1 \leqslant i, j \leqslant 83)$ such that $\left|x_{i}-x_{j}\right| \geqslant 2$, without loss of generality, assume $x_{i}>x_{j}$. Consider a partition of the points of $P$ into groups $Y_{1}, Y_{2}, \cdots, Y_{88}$ (where $Y_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots, 83$), such that
$$
y_{k}=\left|Y_{k}\right|=\left\{\begin{array}{l}
x_{k}, \quad \text { when } k \neq i, \text { and } k \neq j \text {; } \\
x_{i}-1, \text { when } k=i \text {; } \\
x_{j}+1, \text { when } k=j \text {. }
\end{array}\right.
$$

Such a partition exists (just move one point from the original $X_{i}$ to $X_{j}$, and leave the other groups unchanged). For the pattern $G^{\prime}$ obtained from the partition $Y_{1}, Y_{2}, \cdots, Y_{88}$, we have
$$
\begin{array}{l}
m\left(G^{\prime}\right)=C_{y_{1}}^{3}+C_{y_{2}}^{3}+\cdots+C_{y_{y_{3}}}^{3} . \\
\text { Hence } m\left(G^{\prime}\right)-m_{0}=C_{y_{y_{1}}^{3}}^{3}+C_{y_{j}}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{x_{i}-1}^{3}+C_{x_{j}+1}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{x_{x_{i}-1}-1}^{3}+C_{x_{x_{j}}^{2}}^{2}+C_{x_{j}}^{3}-C_{x_{i}-1}^{2}-C_{x_{i}-1}^{3}-C_{x_{j}}^{3} \\
=C_{x_{j}}^{2}-C_{x_{i}-1}^{2} .
\end{array}
$$

Since $x_{j}<x_{i}-1$, we have $\mathrm{C}_{x_{1}}^{2}<\mathrm{C}_{x_{1}-1}^{2}$. Therefore,
$$
m\left(G^{\prime}\right)-m_{0}<0,
$$

which contradicts the minimality of $m_{0}$.
Also, $1994=83 \times 24+2=81 \times 24+2 \times 25$,
so $m_{0}=81 C_{24}^{3}+2 C_{25}^{3}=168544$.
(2) Let the pattern $G^{*}$ be obtained from the groups $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ is the set of points in the $i$-th group $(i=1,2, \cdots, 83)$.
By (1), we can assume
$\left|X_{1}\right|=\left|X_{2}\right|=\cdots=\left|X_{81}\right|=24$,
$\left|X_{82}\right|=\left|X_{83}\right|=25$.
We now provide a coloring method for $G^{*}$ such that $G^{*}$, when colored with 4 different colors, does not contain a triangle with all three sides of the same color.

Let the graph formed by the set $X_{i}$ and its connecting segments be called the $i$-th block of $G^{*}$, denoted as $G_{i}^{*}, i=1,2, \cdots, 83$.
For $G_{83}^{*}$, let
$X_{83}=Y_{1} \cup Y_{2} \cup Y_{3} \cup Y_{4} \cup Y_{5}$,
such that $Y_{\imath} \cap Y_{j}=\varnothing(1 \leqslant i \neq j \leqslant 5)$,
$\left|Y_{i}\right|=5(1 \leqslant i \leqslant 5)$.
Color the segments connecting any two points in each subset $Y_{i}(1 \leqslant i \leqslant 5)$ as shown in Figure 1, and color the segments connecting different subsets $Y_{i}$ and $Y_{j}(i \neq j)$ as shown in Figure 2, where $a, b, c, d$ represent 4 different colors. This way, the colored $G_{83}^{*}$ clearly does not contain a triangle with all three sides of the same color.

For $G_{82}^{*}$, we can use the same coloring method as for $G_{83}^{*}$. As for $G_{i}^{*}(1 \leqslant i \leqslant 81)$, we can first add 1 point and connect it to the original 24 points with a segment each, then color it according to the method for $G_{83}^{*}$, and finally remove the added point and the segments connected to it. This way, the colored $G_{i}^{*}(1 \leqslant i \leqslant 81)$ also does not contain a triangle with all three sides of the same color.
In summary, the conclusion holds.