Given a set of points on a plane $P=\left\{P_{1}, P_{2}, \cdots, P_{1944}\right\}$, 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, thus obtaining a pattern $G$. Different grouping methods result in different patterns. Let the number of triangles in the pattern $G$ with vertices from the 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 the points in $P$) are colored using 4 colors, with each line segment colored exactly one color, prove that there exists a coloring scheme such that $G^{*}$, after coloring, contains no triangle with vertices from the points in $P$ and all three sides of the same color.
Clearly, each pattern is uniquely determined by the grouping method of points in $P$.
(1) Let $m(G)=m_{0}, G$ be obtained by the grouping $X_{1}, X_{2}, \cdots, X_{88}$, 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 $x_{1}+x_{2}+\cdots+x_{83}=1994$, and
$$
m_{0}=C_{X_{1}}^{3}+C_{X_{2}}^{3}+\cdots+C_{X_{83}}^{3} .
$$

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 $x_{i} > x_{j}$, then we can form a new grouping $Y_{1}, Y_{2}, \cdots, Y_{83}$ of points in $P$ such that
$$
y_{k}=\left|Y_{k}\right|=\left\{\begin{array}{l}
x_{k} \quad \text { when } k \neq i \text { and } k \neq j, \\
x_{i}-1 \text { when } k=i, \\
x_{j}+1 \text { when } k=j.
\end{array}\right.
$$

Such a grouping clearly exists. For the pattern $G^{\prime}$ obtained by the grouping $Y_{1}, Y_{2}, \cdots, Y_{83}$, we have
$$
m\left(G^{\prime}\right)=C_{Y_{1}}^{3}+C_{Y_{2}}^{3}+\cdots+C_{\mathrm{Y}_{83}}^{3} .
$$

And
$$
\begin{array}{l} 
m\left(G^{\prime}\right)-m_{0}=C_{X_{i}}^{3}+C_{X_{j}}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{X_{i}-1}^{3}+C_{X_{j}+1}^{3}-C_{X_{i}}^{3}-C_{X_{j}}^{3} \\
=C_{X_{i}-1}^{3}+C_{X_{j}}^{3}+C_{X_{j}}^{2}-C_{X_{i}-1}^{3}-C_{X_{i}-1}^{2}-C_{X_{j}}^{3} \\
=C_{X_{j}}^{2}-C_{X_{i}-1}^{2} . \\
\because x_{j}<x_{i}-1, \quad \therefore C_{X_{j}}^{2}<C_{X_{i}-1}^{2},
\end{array}
$$
$\therefore m\left(G^{\prime}\right)-m_{0}<0$, which contradicts the minimality of $m_{0}$.
$$
\begin{array}{ll}
\because & 1994=83 \times 24+2=81 \times 24+2 \times 25, \\
& m_{0}=81 C_{24}^{3}+2 C_{25}^{3}=168544 .
\end{array}
$$
(2) Let the pattern $G^{*}$ be obtained by the grouping $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ represents 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 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 $Y_{i} \cap Y_{j}=\varnothing(1 \leqslant i \neq j \leqslant 5),\left|Y_{i}\right|=5(1 \leqslant i \leqslant 5)$. We color the segments connecting any two points in each subset $Y_{i}(1 \leqslant i \leqslant 5)$ using the method shown in Figure (1), and we color the segments connecting different subsets $Y_{i}$ and $Y_{j}(i \neq j)$ using the method shown in Figure (2). In the figures, $a, b, c, d$ represent four different colors. The colored $G_{83}^{*}$ clearly does not contain a triangle with all three sides of the same color.

For $G_{82}^{*}$, we can use the same coloring method as for $G_{83}^{*}$. As for $G_{i}^{*}(1 \leqslant i \leqslant 81)$, we can first add a 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. The resulting $G_{i}^{*}$ also does not contain a triangle with all three sides of the same color.
In summary, the conclusion holds.