Example 13 (Problem from the 13th CMO) Find all natural numbers $n$ greater than 3, such that $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}$ divides $2^{2000}$.
Solve: Since 2 is a prime number, this problem is equivalent to finding natural numbers $n>3$, such that $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}=2^{k}(k \in \mathbf{N}_{+}, 1 \leqslant k \leqslant 2000)$, and
$$
\begin{aligned}
1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} & =1+n+\frac{1}{2} n(n+1)+\frac{1}{6} n(n+1)(2 n+1) \\
& =\frac{1}{6}(n+1)\left(n^{2}-n+6\right),
\end{aligned}
$$

Thus, we have $\quad(n+1)\left(n^{2}-n+6\right)=3 \times 2^{k+1}$, making the substitution $m=n+1$, we get
$$
m\left(m^{2}-3 m+8\right)=3 \times 2^{k+1} \text {. }
$$

Therefore, $m \mid 3 \times 2^{k+1}$, so $m$ can only have the following two possible scenarios.
(1) If $m=2^{s}(m>4, s \geqslant 3)$, then $m^{2}-3 m+8=3 \times 2^{t}(t \in \mathbf{N})$.
If $s \geqslant 4$, then $3 \times 2^{t}=\left(2^{s}\right)^{2}-3 \cdot 2^{s}+8 \equiv 8(\bmod 16)$, which implies $t=3$, thus $m^{2}-3 m+8=14$, i.e., $m(m-3)=16$, which is impossible, so only $s=3, m=8$, i.e., $n=7$.
(2) If $m=3 \times 2^{u}(m>4, u \geqslant 1)$, then $m^{2}-3 m+8=2^{v}(v \in \mathbf{N})$.
If $u \geqslant 4$, then $2^{v}=\left(3 \times 2^{u}\right)^{2}-3\left(3 \times 2^{u}\right)+8 \equiv 8(\bmod 16)$,
which implies $v=3$, thus $m^{2}-3 m+8=8$, i.e., $m(m-3)=0$,
which is also impossible. When $u=1$ or 2, i.e., $m=3 \times 2$ or $m=3 \times 2^{2}$, it is impossible for $m^{2}-3 m+8=2^{v}$ to hold.
When $u=3$, i.e., $m=3 \times 2^{3}=24$, then $n=m-1=23$.
On the other hand, when $n=7$, $1+C_{7}^{1}+C_{7}^{2}+C_{7}^{3}=2^{6} \mid 2^{2000}$, and when $n=23$, $1+C_{23}^{1}+C_{23}^{2}+C_{23}^{3}=$ $2^{11} \mid 2^{2000}$.
Therefore, the solutions for $n$ are $n=7$ and $n=23$.