Four. (This question is worth 35 points) Given a set of points on a plane $P=\left\{P_{1}\right.$, $\left.P_{2}, \cdots, P_{1004}\right\}$, where no three points in $P$ are collinear. Divide all the points in $P$ arbitrarily into 83 groups, such that each group has at least 3 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 using 4 different colors, with each segment colored exactly one color, prove that there exists a coloring scheme such that $G^{*}$, after coloring, contains no triangle with all three sides of the same color.
(1) Let $m(G)=m_{0}, G$ be obtained by the partition $X_{1}, X_{2}, \cdots, X_{\text {в3 }}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots$, 83.
$$
\begin{array}{l} 
\text { Let }\left|X_{i}\right|=x_{i}(i=1,2, \cdots, 83) \text {, then } x_{1}+x_{2}+\cdots \\
+x_{83}=1994, \text { and } \\
m_{0}=C_{x_{1}}^{3}+C_{2}^{3}+\cdots+C_{x_{83}}^{3} .
\end{array}
$$

We need to prove that when $1 \leqslant i \neq j \leqslant 83$, we have $\left|x_{i}-x_{j}\right| \leqslant 1$.
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 $x_{i}>x_{j}$, then consider the partition of $P$'s points into $Y_{1}, Y_{2}, \cdots, Y_{83}$ (where $Y_{i}$ is the set of points in the $i$-th group, $1 \leqslant i \leqslant 83$), such that
$$
y_{k}=\left|Y_{k}\right|=\left\{\begin{array}{ll}
x_{k}, & \text { when } k \neq i \text { and } k \neq j, \\
x_{i}-1, & \text { when } k=i, \\
x_{j}+1, & \text { when } k=j.
\end{array}\right.
$$

Such a partition clearly exists. Then, for the pattern $G^{\prime}$ obtained by the partition $Y_{1}, Y_{2}, \cdots, Y_{83}$, we have
$$
m\left(G^{\prime}\right)=C_{y_{1}}^{3}+C_{y_{2}}^{3}+\cdots+C_{y_{83}}^{3} .
$$

And $m\left(G^{\prime}\right)-m_{0}=C_{y_{i}}^{3}+C_{y_{j}}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3}$
$$
\begin{array}{l}
=C_{x_{i}-1}^{3}+C_{x_{j}+1}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{x_{i}-1}^{3}+C_{x_{j}}^{3}+C_{x_{j}}^{2}-C_{x_{i}-1}^{3}-C_{x_{i}-1}^{2}-C_{x_{j}}^{3} \\
=C_{x_{j}}^{2}-C_{x_{i}-1}^{2} . \\
\because x_{j}C_{\left\{x_{j} \mid\right.}^{2}.
\end{array}
$$
\therefore C_{\left|x_{j}\right|-1}^{2}-C_{x_{j} \mid}^{2}>0 \text { . }
$$

This contradicts the minimality of $m_{0}$.
$$
\begin{array}{l}
\because 1994=83 \times 24+2=81 \times 24+2 \times 25, \\
\therefore m_{e}=81 \times C_{24}^{3}+2 \times C_{25}^{3}=168544 .
\end{array}
$$
(2) The same as Solution 1.