Find all natural numbers $n$ greater than 3 such that $1+C_{n}^{1}+C_{n}^{2}+C_{3}^{3}$ divides $2^{2000}$.

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For $n > 3$, we only need to consider which natural numbers $n$ can make $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}=2^{k}(k \in \mathbb{N})$. After calculation, we get
$$
\begin{array}{l}
1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} \\
=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} \\
=\frac{(n+1)\left(n^{2}-n+6\right)}{6},
\end{array}
$$

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

From (1), we must discuss two cases for $m$.
(1) If $m=2^{s}(m>4, s \geqslant 3)$, then
$$
m^{2}-3 m+8=3 \times 2^{t}(t \in \mathbb{N}).
$$

If $s \geqslant 4$, then $3 \times 2^{t} \equiv 8(\bmod 16)$, i.e., $2^{t} \equiv 8 (\bmod 16)$. This implies $2^{t}=8(t=3)$, $m^{2}-3 m+8=24$, i.e., $m(m-3)=16$, which is impossible. Therefore, only $s=3, m=8$, i.e., $n=7$.
(2) If $m=3 \times 2^{u \prime}(m>4, u \geqslant 1)$, then
$$
m^{2}-3 m+8=2^{\nu}(v \in \mathbb{N}).
$$

If $u \geqslant 4$, then $2^{v} \equiv 8(\bmod 16)$. From (1), we know $v=3$, $m(m-3)=0$, which is also impossible. Additionally, when $m=3 \times 2$ or $m=3 \times 2^{2}$, (2) does not hold. Therefore, only $m=3 \times 2^{3}=24$, i.e., $n=23$.

After verification, we find that $1+C_{7}^{1}+C_{7}^{2}+C_{7}^{3}=64=2^{6}$ and $1+C_{23}^{\mathrm{f}}+C_{23}^{2}+C_{23}^{3}=2^{11}$ can both divide $2^{2000}$. Therefore, the answers to this problem are $n=7$ and $n=23$.