4. Let $a_{1}, a_{2}, \cdots, a_{100}$ be
$$
\{1901,1902, \cdots, 2000\}
$$

an arbitrary permutation, and the partial sum sequence
$$
\begin{array}{l}
S_{1}=a_{1}, S_{2}=a_{1}+a_{2}, \\
S_{3}=a_{1}+a_{2}+a_{3}, \\
\cdots \cdots \\
S_{100}=a_{1}+a_{2}+\cdots+a_{100} .
\end{array}
$$

If every term in the sequence $\left\{S_{j}\right\}(1 \leqslant j \leqslant 100)$ cannot be divisible by 3, then the number of such permutations is $\qquad$ kinds.
4. $C_{9}^{33} \times 33!\times 33!\times 34!=\frac{99!\times 33!\times 34!}{66!}$.

Let $\{1901,1902, \cdots, 2000\}=A_{0} \cup A_{1} \cup A_{2}$,
where the elements in $A_{i}$ are congruent to $i \pmod{3}$ for $i=0,1,2$.
Then $\left|A_{0}\right|=\left|A_{1}\right|=33,\left|A_{2}\right|=34$.
Let $a_{i}$ modulo 3 be $a_{i}^{\prime}(i=1,2, \cdots, 100)$. Then whether the partial sums of the permutation $S=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ can be divisible by 3 is determined by the permutation $S^{\prime}=\left\{a_{1}^{\prime}, a_{2}^{\prime}, \cdots, a_{100}^{\prime}\right\}$. Since $S^{\prime}$ contains 33 zeros, 33 ones, and 34 twos, to ensure that no partial sum of $S^{\prime}$ is divisible by 3, the sequence of 67 ones and twos should be
$1,1,2,1,2, \cdots, 1,2$

or $2,2,1,2,1, \cdots, 2,1$.
However, $\left|A_{2}\right|=\left|A_{1}\right|+1$, so only the second case is possible. The 33 zeros, except for $a_{1}^{\prime} \neq 0$, can be placed in any other position, which gives $\mathrm{C}_{9}^{33}$ ways.
Therefore, the number of permutations that meet the requirement is
$$
C_{99}^{33} \times 33!\times 33!\times 34!=\frac{99!\times 33!\times 34!}{66!}(\text { ways }) .
$$