$15 \cdot 4$ 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.
(8th American Invitational Mathematics Examination, 1990)
[Solution]
$$
\begin{aligned}
n! & =n(n-1)(n-2) \cdots 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\
& =24 n(n-1)(n-2) \cdots 5 .
\end{aligned}
$$

At this point, $24 n(n-1)(n-2) \cdots 5$ should be the product of $n-3$ consecutive natural numbers, so we can get $n=23$.
Below we prove that $n=23$ is the largest positive integer $n$ that meets the condition of the problem.
If $n>23$, then
$$
n!=24 n(n-1) \cdots 5n(n-1)(n-2) \cdots 5 \cdot 4 \text {. }
$$

Therefore, we have
$$
n(n-1) \cdots 5 \cdot 4<n!<(n+1) n(n-1) \cdots 5 .
$$

That is, $n!$ lies between the products of two consecutive sets of $n-3$ consecutive natural numbers, so $n!$ cannot be expressed as the product of $n-3$ consecutive natural numbers.
When $n=23$, it is clear that
$$
23!=24 \cdot 23 \cdot 22 \cdot \cdots \cdot 6 \cdot 5 \text {. }
$$

Therefore, the largest positive integer $n$ sought is 23.