Example 6 Find the largest positive integer $n$ that satisfies the following condition: $n$ is divisible by all positive integers less than $\sqrt[3]{n}$. (1998, Asia Pacific Mathematical Olympiad)
Analysis and Solution: The problem involves radicals, which are inconvenient to handle. If we set the largest integer less than $\sqrt[3]{n}$ as $k$, then the original problem can be transformed into the following equivalent form:
Let $k$ and $n$ be positive integers, satisfying
$$
k^{3} < n < (k+1)^{3},
$$

i.e., it holds for integers $k \geqslant 9$.
Thus, when $k \geqslant 9$, $n$ lies between 1 and 2 times $k(k-1)(k-2)$, which cannot satisfy the given conditions.

Upon calculation, when $k=7$, the least common multiple (LCM) of 1 to $k$ is 420, which lies between $7^{3}=343$ and $8^{3}=512$. Therefore, $n=420$ is the largest solution when $k$ is odd.

When $k$ is even, $k-1, k-2, k-3$ are pairwise coprime, so $(k-1)(k-2)(k-3)$ divides $n$. In this case, we still have
$$
(k-1)(k-2)(k-3) < k^{3} < n.
$$

The inequality
$$
(k+1)^{3} < 2(k-1)(k-2)(k-3)
$$

holds for positive integers $k \geqslant 14$. Therefore, the allowed values of $k$ must be even numbers not exceeding 12.

Upon inspection, when $k=8, 10, 12$, the LCM of the positive integers from 1 to $k$ is greater than $(k+1)^{3}$, which does not meet the given conditions; and for $k \leqslant 6$, the corresponding $n$ must not exceed $7^{3} = 343$.
In summary, the answer to this problem is 420.
Note: In this problem, we used the conclusion: "For positive integers $a$ and $b$, if there exists an integer $n$ such that $na < b < (n+1)a$, then $a$ does not divide $b$."