Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest positive integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive positive integers.

The product of $n - 3$ consecutive integers can be written as $\frac{(n - 3 + a)!}{a!}$ for some integer $a$. Thus, $n! = \frac{(n - 3 + a)!}{a!}$, from which it becomes evident that $a \ge 3$. Since $(n - 3 + a)! > n!$, we can rewrite this as $\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!$. For $a = 4$, we get $n + 1 = 4!$ so $n = 23$. For greater values of $a$, we need to find the product of $a-3$ consecutive integers that equals $a!$. $n$ can be approximated as $^{a-3}\sqrt{a!}$, which decreases as $a$ increases. Thus, $n = \boxed{23}$ is the greatest possible value to satisfy the given conditions.