Example 3 Let $a_{1}, a_{2}, \cdots, a_{100}$ be any permutation of $\{1901,1902, \cdots, 2000\}$, and the partial sum sequence $S_{1}=a_{1}$, $S_{2}=a_{1}+a_{2}, S_{3}=a_{1}+a_{2}+a_{3}, \cdots, S_{100}=a_{1}+a_{2}+\cdots+a_{100}$. If every term in the sequence $S_{j}(1 \leqslant j \leqslant 100)$ is not divisible by 3, how many such sequences are there?
(2000 Canadian Olympiad Problem)
Let $\{1901,1902, \cdots, 2000\}=R_{0} \cup R_{1} \cup R_{2}$, where any element in $R_{i}$ is congruent to $i$ modulo 3 $(i=0,1,2)$, then $\left|R_{0}\right|=\left|R_{1}\right|=33,\left|R_{2}\right|=34$. Let $a_{i}$ modulo 3 be $a^{\prime}{ }_{i}(i=1,2, \cdots, a^{\prime}{ }_{100})$ which determines $S^{\prime}$. In $S^{\prime}$, there are 33 0s, 33 1s, and 34 2s. To ensure that no partial sum of $S$ is divisible by 3, the 67 numbers in $S^{\prime}$ composed of 1s and 2s should be $1,1,2,1,2, \cdots, 1,2$ or $2,2,1,2,1, \cdots, 2,1$. However, $\left|R_{2}\right|=\left|R_{1}\right|+1$, so only the second scenario is possible. In $S^{\prime}$, the 33 0s, except for $a^{\prime}{ }_{1} \neq 0$ which can be placed in any other position, can be arranged in $C_{99}^{33}$ ways. Therefore, the number of permutations that satisfy the condition is $C_{99}^{33} \cdot 33!33!34!=\frac{99!33!34!}{66!}$.