Example 10 We notice that $6!=8 \cdot 9 \cdot 10$. Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers.
Solution: According to the requirements of the problem, write $n!$ as
$$
n!=n(n-1) \cdot \cdots \cdot 5 \cdot 24 \text {. }
$$

Therefore, $n+1 \leqslant 24, n \leqslant 23$. Hence, the maximum value of $n$ is 23. At this point, we have
$$
23!=24 \times 23 \times \cdots \times 5 .
$$

The right-hand side of the above equation is the product of 20 consecutive natural numbers.