(4) Find all natural numbers $n$ greater than 3 such that $1+\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}$ divides $2^{2000}$.
Solve: Since 2 is a prime number, we only need to consider which natural numbers $n>3$ can make
$$
1+\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}=2^{k}, k \in \mathbf{N}
$$

After calculation, we get
$$
\begin{array}{l}
1+\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}= \\
1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}= \\
\frac{(n+1)\left(n^{2}-n+6\right)}{6}
\end{array}
$$

That is
$$
(n+1)\left(n^{2}-n+6\right)=3 \times 2^{k+1}
$$

By making the substitution $m=n+1$, we get
$$
m\left(m^{2}-3 m+8\right)=3 \times 2^{k+1}
$$

From (1), we must discuss $m$ in two cases.
i 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 $3 \times 2^{t} \equiv 8(\bmod 16)$, 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., $n=7$.
ii 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 $2^{v} \equiv 8(\bmod 16)$. From $\mathrm{i}$, 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 $1+\mathrm{C}_{7}^{1}+\mathrm{C}_{7}^{2}+\mathrm{C}_{7}^{3}=64=2^{6}$ and $1+\mathrm{C}_{23}^{1}+\mathrm{C}_{23}^{2}+\mathrm{C}_{23}^{3}=2^{11}$ can both divide $2^{2000}$. Therefore, the answer to this problem is $n=7$ and $n=23$.