$2 \cdot 91$ 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.

Expressing the above text in English, while retaining the original text's line breaks and format, results in the output above.
[Solution] Since
$$
\begin{aligned}
n! & =n(n-1) \cdots 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\
& =24 \cdot n \cdot(n-1) \cdots 5 .
\end{aligned}
$$

When $n=23$, $24 \cdot n \cdot(n-1) \cdots 5$ is the product of $n-3$ consecutive natural numbers.
Below we prove that $n=23$ is the largest positive integer $n$ that meets the condition.
If $n>23$, then by
$$
\begin{aligned}
n! & =k \cdot n \cdot(n-1) \cdots(n-t) \\
& =n \cdot(n-1) \cdots(n-t) k,
\end{aligned}
$$

we know that $k>n$, thus $k \geqslant 25$.
Therefore, $k$ is at least the product of the 5 smallest consecutive natural numbers, and $n!$ can only be expressed as the product of fewer than $n-3$ consecutive natural numbers, which contradicts the problem statement.
Thus, the largest positive integer $n$ that meets the condition is 23.