8 (CMO-13 Problem) Find all natural numbers greater than 3, such that $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}$ divides $2^{2000}$.
8. Since 2 is a prime number, this problem is equivalent to finding natural numbers $n>3$, such that
$1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}=2^{k} \quad(k \in \mathbf{N}, k \leqslant 2000)$.
$1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}=\frac{(n+1)\left(n^{2}-n+6\right)}{6}$,
thus we have $(n+1)\left(n^{2}-n+6\right)=3 \times 2^{k+1}$.
By making the substitution $m=n+1$, we get
$$
m\left(m^{2}-3 m+8\right)=3 \times 2^{k+1} \text {. }
$$

We now classify and discuss $m$.
(1) If $m=2^{x} \quad(m>4, s \geqslant 3)$, then $m^{2}-3 m+8=3 \times 2^{t} \quad(t \in \mathbf{N})$.
If $s \geqslant 4$, then $m^{2}-3 m+8=3 \times 2^{t} \equiv 8(\bmod 16)$.
From this, we know that $t=3$, thus
$m^{2}-3 m+8=24$, which means $m(m-3)=16$. This is impossible.
Therefore, only $s=3, m=8$, i.e., $n=7$.
(2) If $m=3 \times 2^{u} \quad(m>4, u \geqslant 1)$, then $m^{2}-3 m+8=2^{v} \quad(v \in \mathbf{N})$.
If $u \geqslant 4$, then $m^{2}-3 m+8=2^{v} \equiv 8(\bmod 16)$.
From this, we know 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 $1+C_{7}^{1}+C_{7}^{2}+C_{7}^{3}=64=2^{6} \mid 2^{2000}$, and $1+C_{23}^{1}+C_{23}^{2}+C_{23}^{3}=2^{11} \mid 2^{2000}$.
Therefore, the solutions to this problem are $n=7$ and $n=23$.