There are 1994 points on a circle, which are painted in several different colors, and the number of points of each color is different. Now, take one point from each color set to form a polygon with vertices of different colors inside the circle. To maximize the number of such polygons, how many different colors should the 1994 points be painted, and how many points should each color set contain?
Let 1994 points be colored with $n$ colors, and the number of points of each color, in ascending order, is $m_{1}1$. In fact, if $m_{1}=1$, because $m_{1} m_{2} \cdots m_{n}m_{k+1} m_{k}$.
Therefore, when $m^{\prime}{ }_{n+1}, m^{\prime} \star$ replace $m_{\star+1}, m_{\star}$, the value of $M$ increases, leading to a contradiction:
(2) $m_{i+1}-m_{i}=2$ holds for exactly one $i$. For this:
(i) $m_{i+1}-m_{i}=2$ holds for at most one $i$. If otherwise, there exist positive integers $j, k$ such that $1 \leqslant j < k \leq n$ and $m_{j+1} - m_{j} = m_{k+1} - m_{k} = 2$. Then, $m_{j+1} = m_{j} + 2$ and $m_{k+1} = m_{k} + 2$. If we replace $m_{k+1}, m_{j}$ with $m^{\prime}{ }_{k+1}, m^{\prime} j_{j}$, the value of $M$ increases, leading to a contradiction.
(ii) If $m_{i+1}-m_{i}=1$ for all $i=1,2, \cdots, n-1$, then $\sum_{i=1}^{n} m_{i} = n m_{1} + \frac{1}{2} n(n-1) = 1994 = 2 \times 997 \Rightarrow n(2 m_{1} - 1 + n) = 4 \times 997$. Since $n$ and $2 m_{1} - 1 + n$ are one odd and one even, and $2 m_{1} - 1 + n > n$, and 997 is a prime number, the only solution is $n=4$, $2 m_{1} - 1 + n = 997$, giving $m_{1} = 497 = 495 + 22$. If we take $m^{\prime} = m_{k+2} - 2$, then $m_{k} < 2$ and $m_{k+2} > 4$. Hence, $2 m^{\prime} > m_{k+2}$. Replacing $m_{k+2}$ with 2 and $m^{\prime}$, the value of $M$ increases, leading to a contradiction. Therefore, $m_{1} = 2$.
(II) From (I), we can assume the number of points of each color is 2, 3, ..., $k-1$, $k+1$, $k+2$, ..., $n$, $n+1$, $n+2$ ($k \leqslant n-1$).
We have $\frac{1}{2}(n+1)(n+4) - k = 1994$.
This gives $\left\{\begin{array}{l}n^{2} + 5n - 3984 > 0, \\ n^{2} + 3n - 3982 \leqslant 0 .\end{array}\right.$

Solving, we get $61 \leqslant n < 62$.
Taking $n=61$, we have $k = \frac{1}{2}(n+1)(n+4) - 1994 = 31 \times 65 - 1994 = 21$. Therefore, 1994 points can be colored with 61 colors, with the number of points of each color being $2, 3, \cdots, 19, 20, 22, 23, \cdots, 61, 62, 63$. In this case, the resulting polygon is a 61-sided polygon, and the number of such polygons is the maximum. (Xu Jingling, Jiugu Middle School, Susong County, Anhui Province)