1.91 Find all natural numbers greater than 3, such that $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}$ divides $2^{2000}$.
(13th China High School Mathematics Winter Camp, 1998)
[Solution] Since 2 is a prime number, this problem is equivalent to finding natural numbers $n>3$ such that
$$
\begin{aligned}
1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} & =2^{k}(k \in \mathbb{N}, k \leqslant 2000). \\
1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} & =1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} \\
& =\frac{(n+1)\left(n^{2}-n+6\right)}{6}
\end{aligned}
$$

Thus, we have $(n+1)\left(n^{2}-n+6\right)=3 \times 2^{k+1}$.
By substituting $m=n+1$, we get
$$
m\left(m^{2}-3 m+8\right)=3 \times 2^{k+1}.
$$

We now classify and discuss $m$.
(1) If $m=2^{c}(m>4, s \geqslant 3)$, then
$$
m^{2}-3 m+8=3 \times 2^{t}(t \in \mathbb{N}).
$$

If $s \geqslant 4$, then $m^{2}-3 m+8=3 \times 2^{t} \equiv 8(\bmod 16)$.
This implies that $t=3$, thus
$$
m^{2}-3 m+8=24, \text{ i.e., } m(m-3)=16.
$$

This is impossible.
Therefore, only $s=3, m=8$, i.e., $n=7$.
(2) If $m=3 \times 2^{u}(m>4, u \geqslant 1)$, then
$$
m^{2}-3 m+8=2^{v}(v \in \mathbb{N}).
$$

If $u \geqslant 4$, then
$$
m^{2}-3 m+8=2^{v} \equiv 8(\bmod 16),
$$

This implies that $v=3$. Hence $m(m-3)=0$. This is also impossible.
For $u=1$ and $u=2$, i.e., $m=3 \times 2$ and $m=3 \times 2^{2}$, neither can satisfy $m^{2}-3 m+8=2^{v}$.
When $m=3 \times 2^{3}=24$, we can find $n=23$.
Since $\quad 1+C_{7}^{1}+C_{7}^{2}+C_{7}^{3}=64=2^{6} \leqslant 2^{2000}$,
and $1+C_{23}^{1}+C_{23}^{2}+C_{23}^{3}=2^{11} \leqslant 2^{2000}$.
Therefore, the solutions to this problem are $n=7$ and $n=23$.