$4 \cdot 29$ Given a set of points $P=\left\{P_{1}, P_{2}, \cdots, P_{1994}\right\}$ in the plane, where no three points are collinear. Divide all points in $P$ into 83 groups arbitrarily, such that each group has at least 3 points, and each point belongs to exactly one group. Then connect every pair of points within the same group with a line segment, and do not connect points in different groups, thus obtaining a graph $G$. Different grouping methods result in different graphs. Let $m(G)$ denote the number of triangles in graph $G$ with vertices from $P$.
(1) Find the minimum value $m_{0}$ of $m(G)$;
(2) Let $G^{*}$ be a graph such that $m\left(G^{*}\right)=m_{0}$, and color the edges (referring to line segments with endpoints in $P$) of $G^{*}$ using 4 colors, with each edge colored exactly one of the four colors. Prove that there exists a coloring scheme such that $G^{*}$, after being colored, contains no monochromatic triangle (i.e., a triangle with all three sides the same color).
[Solution] Since there are only a finite number of ways to divide the point set $P$ into 83 groups, there must be a grouping method that minimizes the number of triangles formed. Let the corresponding graph be $G$, then $m(G)=m_{0}$. Suppose the graph $G$ is obtained by the grouping $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ represents the set of points in the $i$-th group, and let $x_{i}=\left|X_{i}\right|, i=1,2, \cdots, 83$. Thus, we have $x_{1}+x_{2}+\cdots+x_{83}=1994$, and
$$
m_{0}=C_{x_{1}}^{3}+C_{x_{2}}^{3}+\cdots+C_{x_{83}}^{3}.
$$

If there exist $x_{i}, x_{j}$ such that $\left|x_{i}-x_{j}\right| \geqslant 2 (1 \leqslant i < j \leqslant 83)$, and without loss of generality, assume $x_{i} > x_{j}$, then when one point from $X_{i}$ is moved to $X_{j}$ while all other points remain unchanged, let the graph corresponding to the resulting grouping be $G^{\prime}$, and $m\left(G^{\prime}\right)=m_{1}$. Thus, we have
$$
\begin{aligned}
m_{0}-m_{1} & =C_{x_{i}}^{3}+C_{x_{j}}^{3}-C_{x_{i}-1}^{3}-C_{x_{j}+1}^{3} \\
& =\frac{1}{2}\left[\left(x_{i}-1\right)\left(x_{i}-2\right)-x_{j}\left(x_{j}-1\right)\right]>0,
\end{aligned}
$$

which contradicts the minimality of $m_{0}$. Therefore, for any $1 \leqslant i < j \leqslant 83$, we have $\left|x_{i}-x_{j}\right| \leqslant 1$.

Since $1994=83 \times 24+2=81 \times 24+2 \times 25$, there must be 81 groups with 24 points each and 2 groups with 25 points each. Hence,
$$
m_{0}=81 C_{24}^{3}+2 C_{25}^{3}=168544.
$$

Let the grouping corresponding to $G^{*}$ be $X_{1}, X_{2}, \cdots, X_{83}$, and $x_{1}=x_{2}=25, x_{3}=x_{4}=\cdots=x_{83}=24$. Below, we provide a coloring method for $G^{*}$ such that after coloring $G^{*}$ with 4 colors, there are no monochromatic triangles.

Since there are no edges between different groups, it is sufficient to color the edges within each group. Since each group has either 25 points or 24 points, it is only necessary to provide a coloring method that meets the requirements for a group of 25 points, and for a group of 24 points, just remove one point from the former.
Divide the 25 points in $X_{1}$ into 5 subsets:
$$
\begin{array}{l}
X_{1}=M_{1} \cup M_{2} \cup M_{3} \cup M_{4} \cup M_{5}, \\
\left|M_{i}\right|=5, i=1,2,3,4,5.
\end{array}
$$

Let the graph formed by the points and edges in $X_{1}$ be denoted as $G_{1}^{*}$, and the graph formed by the points and edges in $M_{i}$ be denoted as $G_{1i}^{*}$. Let $a, b, c, d$ represent 4 different colors. Color $G_{1i}^{*}$ as shown in the following figure (a), and then color the edges between the 5 subgraphs in $G_{1}^{*}$ as shown in the following figure (b). It is easy to see that in this coloring method, there are no monochromatic triangles in the colored $G_{1}^{*}$.