Example 8 There are 1996 points on a circle, which are colored with several different colors. Now, from each set of points of the same color, one point is taken to form an inscribed polygon with vertices of different colors. If the number of points of each color is different, to maximize the number of such polygons, how many different colors should the 1996 points be colored with? And how many points should be of each color?
Solution: Let 1996 points be colored with $n$ colors, where each point is colored one of the $n$ colors, and let the number of points of each color be $m_{1}, m_{2}, m_{3}, m_{4}, \cdots, m_{n}$. Without loss of generality, assume $m_{1} \geq 1$, and $m_{i} - m_{i-1} \leq 2 (i=2,3, \cdots, n)$, otherwise, if $m_{i} - m_{i-1} \geq 3$, take $m^{\prime}_{i-1} = m_{i-1} + 1, m^{\prime}_{i} = m_{i} - 1$, then $m^{\prime}_{i-1} \cdot m^{\prime}_{i} = (m_{i-1} + 1)(m_{i} - 1) = m_{i-1} \cdot m_{i} + m_{i} - m_{i-1} - 1 > m_{i-1} \cdot m_{i}$. This implies that $S$ cannot be maximized.

Also, $m_{i} - m_{i-1} = 2$ can occur at most once, otherwise, if $m_{i} - m_{i-1} = 2, m_{j} - m_{j-1} = 2 (1 < i < j \leq n)$, then $m_{i-1} + 2 + m_{j-1} = m_{i} + m_{j}$, but $m_{i-1} \cdot 2 \cdot m_{j-1} > m_{i} \cdot m_{j}$, which means such an $S$ cannot be maximized.

We have $m_{1} \leq 3$, otherwise, if $m_{1} \geq 5$, then $2 + (m_{1} - 2) = m_{1}$, but $2(m_{1} - 2) = m_{1} + (m_{1} - 4) > m_{1}$, thus $S' = 2(m_{1} - 2) \cdot m_{2} \cdot m_{3} \cdots \cdot m_{n} > m_{2} \cdot m_{1} \cdots \cdot m_{n} = S$, i.e., $S$ is not maximized. If $m_{1} = 4$, then $m_{2} = 5$ or 6. When $m_{1} = 4, m_{2} = 5$, $2 + 3 + 4 = m_{1} + m_{2}$, but $2 \times 3 \times 4 > m_{1} \cdot m_{2}$; when $m_{1} = 4, m_{2} = 6$, $2 + 3 + 5 = m_{1} + m_{2}$, but $2 \times 3 \times 5 > m_{1} \cdot m_{2}$; therefore, when $m_{1} \geq 4, S$ cannot be maximized, i.e., $2 \leq m_{1} \leq 3$.

The sum of $n$ consecutive numbers starting with 2 or 3 cannot be 1996. Otherwise, if the first term is 2, then $\frac{1}{2} \cdot n \cdot (n + 3) = 1996$, i.e., $n(n + 3) = 8 \times 499$, and $n$ and $n + 3$ have different parities and $n < 499$. If the first term is 3, then $m_{1} + m_{2} + \cdots + m_{i-1} + m_{i} + \cdots + m_{n} = 2 + m_{1} + m_{2} + \cdots + m_{i-1} + [m_{i+1} - 2] + m_{i} + m_{i+2} + m_{i+3} + \cdots + m_{n}$, since $2(m_{i+1} - 2) = m_{i+1} + (m_{i+1} - 4) > m_{i+1}$, then $2 m_{1} \cdot m_{2} \cdots \cdot m_{i-1} [m_{i+1} - 2] \cdot m_{i} \cdot m_{i+2} \cdots \cdot m_{n} > m_{1} \cdot m_{2} \cdots \cdot m_{n}$, i.e., if $m_{1} = 3, S$ cannot be maximized. Therefore, to maximize $S$, we must have $m_{1} = 2$, and $m_{i} - m_{i-1} = 2$ occurs exactly once, and the rest $m_{j} - m_{j-1} = 1$, then we can set $m_{1}, m_{2}, \cdots, m_{n}$ as $2, 3, 4, \cdots, i-1, i+1, i+2, \cdots, n+1, n+2$, their sum is $\frac{(1+n)(n+4)}{2} - i$, thus $(1+n)(n+4) = 2 \times 1996 + 2i (i < n)$, solving this gives $n = 61, i = 19$. Therefore, 1996 points can be colored with 61 different colors, the number of points of each color being $2, 3, 4, \cdots, 18, 20, 21, \cdots, 61, 62, 63$, at this point the polygon is a 61-sided polygon, and the number is maximized.