Example 1 Let the set of points on a plane be $P=\left\{P_{1}, P_{2}, \cdots, P_{194}\right\},$ where no three points in $P$ 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 any two points within the same group with a line segment, and do not connect any two points from different groups, thus forming a pattern $G$. Different grouping methods result in different patterns, and the number of triangles in the pattern with vertices from points in $P$ is 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 four colors, with each line segment colored exactly one color, prove: there exists a coloring scheme such that after coloring $G^{*}$, there is no triangle with vertices from points in $P$ and all three sides of the same color.
(1) Since the number of grouping methods is finite, there exists a pattern $G$ such that $m(G)=m_{0}$. Suppose pattern $G$ satisfies $m(G)=$ $m_{0}, G$ is composed of groups $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ is the set of points in the $i$-th group $(i=1,2, \cdots, 83)$, and let $\left|X_{i}\right|=x_{i}$, then $x_{1}+x_{2}+\cdots+x_{83}=1994$,
$$
m_{0}=\sum_{i=1}^{83} C_{x_{i}}^{3} .
$$

We now prove: For any $1 \leqslant i < j \leqslant 83$, if $x_{i} \geqslant x_{j}+2$, then we can construct a new pattern $G^{\prime}$ by moving one point from $X_{i}$ to $X_{j}$. Since $x_{i} \geqslant x_{j}+2$, we have $x_{i}-1 \geqslant x_{j}+1$, so $m\left(G^{\prime}\right)-m(G)=C_{x_{j}}^{2}-C_{x_{i}-1}^{2}<0$, i.e., $m\left(G^{\prime}\right)<m(G)=$ $m_{0}$, which contradicts the fact that $m_{0}$ is the minimum value.

Since $1994=83 \times 24+2=81 \times 24+2 \times 25$, when $m(G)=m_{0}$, $x_{1}, x_{2}, \cdots, x_{83}$ include 81 values equal to 24 and 2 values equal to 25. Therefore, $m_{0}=81 C_{24}^{3}+2 C_{25}^{3}=168544$.
(2) From (1), we know that for a pattern $G^{*}$ such that $m\left(G^{*}\right)=m_{0}$, $G^{*}$ can be divided into 83 mutually unconnected sub-patterns $G_{1}^{*}, G_{2}^{*}, \cdots, G_{83}^{*}$, where 81 sub-patterns $G_{1}^{*}, G_{2}^{*}, \cdots, G_{81}^{*}$ contain 24 points, and two sub-patterns $G_{82}^{*}, G_{83}^{*}$ contain 25 points, and let the point set of $G_{i}^{*}$ be $X_{i}(i=1,2, \cdots, 83)$. For $G_{83}^{*}$, let $X_{83}=Y_{1} \cup Y_{2} \cup \cdots \cup$ $Y_{5}, Y_{i} \cap Y_{j}=\varnothing(1 \leqslant i<j \leqslant 5)$ and $\left|Y_{i}\right|=5(1 \leqslant i \leqslant 5)$, and each $Y_{i}$ is colored using the method shown in Figure 4-10. The line segments connecting different $Y_{i}$ and $Y_{j}$ are colored using the method shown in Figure 4-11, where $a, b, c, d$ represent 4 different colors. The colored $G_{83}^{*}$ clearly does not contain a triangle with vertices from $P$ and all three sides of the same color.

For $G_{82}^{*}$, the same coloring method as $G_{83}^{*}$ can be used. For $G_{i}^{*}(1 \leqslant i \leqslant 81)$, first add one point and connect it to all 24 existing points, then color it according to the method for $G_{83}^{*}$, and finally remove the added point and the line segments connected to it. The resulting colored $G_{i}^{*}$ clearly does not contain a triangle with vertices from $P$ and all three sides of the same color.
In summary, the conclusion holds.