Four. (This question is worth 35 points) Given a set of points on a plane $P=\left\{P_{1}, P_{2}, \cdots, P_{1994}\right\}$, where no three points in $P$ are collinear. Divide all the 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 any two points in the same group with a line segment, and do not connect points in different groups. This results in 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 in $G^{*}$ (referring to line segments with endpoints from points in $P$) are colored with 4 different colors, and each line segment is colored with exactly one color, prove that there exists a coloring scheme such that $G^{*}$, after coloring, does not contain a triangle with all three sides of the same color.
Certainly, here is the translation of the provided text into English, preserving the original formatting:

---

Clearly, each pattern is uniquely determined by the grouping method of points in $P$.
(1) Let $m(G)=m_{0}, G$ be obtained from the grouping $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1, 2, \cdots, 83$, and let $\left|X_{i}\right|=x_{i}(i=1,2, \cdots, 83)$. Then we have $x_{1}+x_{2}+\cdots+x_{83}=1994$, and $m_{0}=\mathrm{C}_{x_{1}}^{3}+\mathrm{C}_{x_{2}}^{3}+\cdots+\mathrm{C}_{x_{83}}^{3}$.
We now 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 $x_{1}>x_{2}$. Then consider the grouping of points in $P$ as $Y_{1}, Y_{2}, \cdots, Y_{83}$, where $Y_{i}$ is the set of points in the $i$-th group, $1 \leqslant i \leqslant 83$, such that
$$
y_{k}=\left|Y_{k}\right|=\left\{\begin{array}{l}
x_{k} \text { when } k \neq i \text { and } k \neq j \\
x_{i}-1 \text { when } k=i \\
x_{i}+1 \text { when } k=j
\end{array}\right.
$$

Such a grouping clearly exists. For the pattern $G^{\prime}$ obtained from the grouping $Y_{1}, Y_{2}, \cdots, Y_{83}$, we have
$$
\begin{array}{l}
m\left(G^{\prime}\right)=\mathrm{C}_{Y_{1}}^{3}+\mathrm{C}_{Y_{2}}^{3}+\cdots+\mathrm{C}_{Y_{83}}^{3} . \\
\text { and } m\left(G^{\prime}\right)-m_{0}=\mathrm{C}_{y_{i}}^{3}+\mathrm{C}_{y_{j}}^{3}-\mathrm{C}_{x_{i}}^{3}-\mathrm{C}_{x_{j}}^{3} \\
=\mathrm{C}_{x_{i}-1}^{3}+\mathrm{C}_{x_{j}+1}^{3}-\mathrm{C}_{x_{i}}^{3}-\mathrm{C}_{x_{j}}^{3} \\
=\mathrm{C}_{x_{i}-1}^{3}+\mathrm{C}_{x_{j}}^{3}+\mathrm{C}_{x_{j}}^{2}-\mathrm{C}_{x_{i}-1}^{3}-\mathrm{C}_{x_{i}-1}^{2}-\mathrm{C}_{x_{j}}^{3} \\
=\mathrm{C}_{x_{j}}^{2}-\mathrm{C}_{x_{i}-1}^{2}, \\
\because x_{j}<x_{i}-1, \therefore \mathrm{C}_{x_{j}}^{2}<\mathrm{C}_{x_{i}-1}^{2},
\end{array}
$$
$\therefore m\left(G^{\prime}\right)-m_{0}<0$, which contradicts the minimality of $m_{0}$.
$$
\begin{array}{l}
\because 1994=83 \times 24+2=81 \times 24+2 \times 25, \\
\therefore m_{0}=81 \mathrm{C}_{24}^{3}+2 \mathrm{C}_{25}^{3}=168544 .
\end{array}
$$
(2) Let the pattern $G^{*}$ be obtained from the grouping $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ is the set of points in the $i$-th group $(i=1,2, \cdots, 83)$. By (1), 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 four different colors, does not contain a triangle with all three sides of the same color.
We call the graph formed by the set $X_{i}$ and its connecting segments 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 $Y_{i} \cap Y_{j}=\Phi(1 \leqslant i \leqslant 5),\left|Y_{i}\right|=5(1 \leqslant i \leqslant 5)$. Color each subset of segments as shown in Figure (2), where $a, b, c, d$ represent four different colors. The colored $G_{83}^{*}$ clearly does not contain a triangle with all three sides of the same color.
For $G_{82}^{*}$, use the same coloring method as for $G_{83}^{*}$. For $G_{i}^{*}(1 \leqslant i \leqslant 81)$, add one point and connect it to the original 24 points with one segment each, then color it according to the method for $G_{83}^{*}$, and finally remove the added point and the segments connected to it. The resulting $G_{i}^{*}$ also does not contain a triangle with all three sides of the same color. Therefore, the conclusion holds.

---