10 days * * 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.


Parse $n!=n(n-1) \cdots 5 \cdot 4!$. Therefore $n+1 \leqslant 4!=24$, so $n=23$ is the desired value. At this point, $23!=24 \times 23 \times \cdots \times 6 \times 5$.