5. Notice that $6!=8 \times 9 \times 10$. Then, the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers is $\qquad$ .
5.23.

According to the requirements of the problem, write $n!$ as
$$
n!=n \times(n-1) \times \cdots \times 5 \times 24 \text {. }
$$

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

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