[Example 3.5.4] Find all natural numbers greater than 3 such that
$$
1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3} \mid 2^{2000} .
$$
First, adopt a generalization strategy: Since 2 is a prime number, we only need to consider which natural numbers $n>3$ make $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}$ have the shape of $2^{k}$, i.e.,
find $n$ such that $1+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}=2^{k}, n>3, n \in \mathbf{N}$.
That is, $1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}=2^{k}$,
rearranging gives
$$
(n+1)\left(n^{2}-n+6\right)=3 \times 2^{k+1} .
$$

Make a substitution $m=n+1$, then the above equation becomes
$$
m\left(m^{2}-3 m+8\right)=3 \times 2^{k+1} .
$$

Since 2 and 3 are both prime numbers, it is clear that $m$ can only have the form $2^{s}$ or $3 \times 2^{t}$. Therefore, the problem must be divided into two sub-problems.
(1) When $m=2^{s}(s \geqslant 3)$, at this time (1) becomes
$$
m^{2}-3 m+8=3 \times 2^{u} \text {, }
$$

If $s \geqslant 4$, at this time
$$
m^{2}-3 m+8 \equiv 8 \quad(\bmod 16),
$$

thus
$$
3 \times 2^{u} \equiv 8 \quad(\bmod 16),
$$

i.e.,
$$
2^{u} \equiv 8 \quad(\bmod 16),
$$

so $u=3$, i.e.,
$$
\begin{array}{l}
m^{2}-3 m+8=3 \times 2^{3}=24 \\
m(m-3)=16
\end{array}
$$

There are no integer solutions at this time.
Therefore, only $s=3$, i.e.,
$$
m=2^{3}=8, n=m-1=7 .
$$
(2) When $m=3 \times 2^{t}(t \geqslant 1)$, then (1) becomes
$$
m^{2}-3 m+8=2^{v} \text {. }
$$

Similarly, if $t \geqslant 4$, then
$$
\begin{array}{l}
m^{2}-3 m+8 \equiv 8 \quad(\bmod 16) . \\
2^{v} \equiv 8 \quad(\bmod 16),
\end{array}
$$

at this time $v=3$, i.e., $m^{2}-3 m+8=8$, solving gives $m=0$ or 3, which contradicts the condition $n>3, m>4$.
Therefore, $t \leqslant 3$.
When $t=1,2,3$, $m$ is correspondingly $6,12,24$.
Substituting $m=6,12$ into (2) gives
$$
\begin{array}{l}
36-18+8=26 \neq 2^{v}, \\
144-36+8=116 \neq 2^{v} .
\end{array}
$$

Substituting $m=24$ into (2) gives
$$
576-72+8=512=2^{9} \text {. }
$$

Thus,
$$
n=23 \text {. }
$$

Upon verification, for $n=7$ and $n=23$, we have
$$
\begin{array}{c}
1+C_{7}^{1}+C_{7}^{2}+C_{7}^{3}=64=2^{6} \mid 2^{2000}, \\
1+C_{23}^{1}+C_{23}^{2}+C_{23}^{3}=2048=2^{11} \mid 2^{2000} .
\end{array}
$$

Therefore, $n=7, n=23$ are the solutions.