11. 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.
11. If $n!$ can be expressed as the product of $(n-3)$ consecutive integers, then there exists an integer $k$, such that
$$
\begin{array}{l} 
n!=(n+k)(n+k-1) \cdots(k+4) \\
=-(n+k)! \\
(k+3)!
\end{array}
$$

Originally,
$$
\begin{array}{l}
\frac{n+k}{k+3} \cdot \frac{n+k-1}{k+2} \cdot \ldots \cdot \frac{n+2}{5} \cdot \frac{n+1}{4!} \\
=1
\end{array}
$$

It is evident that $n>23$ is impossible.
On the other hand, $n=23, k=1$, is clearly one of its solutions. Therefore, the value of $n$ that satisfies the problem's requirements is 23.