The answer is $420$.

Let us consider the positive integer $m$ so that $m^3 \leq n < (m+1)^3$. As $n = 420$ satisfies the conditions, we will consider the case when $m \geq 7$. Note that each of $m$, $m-1$, $m-2$ and $m-3$ divides $n$ and hence $\operatorname{lcm}(m, m-1, m-2, m-3)$ divides $n$. Since $\gcd(n-1, n-2) = 1$, $\gcd(n, n-3)$ divides $3$ and $\gcd(n(n-3), (n-1)(n-2))$ divides $2$, we have that

$$\frac{m(m-1)(m-2)(m-3)}{6}$$
divides $n$. Therefore,
$$\frac{m(m-1)(m-2)(m-3)}{6} < (m+1)^3$$
and hence $m \leq 12$.

If $m = 11$ or $12$, then $11 \cdot 7 \cdot 5 \cdot 9 \cdot 8 = 27720 \mid n$, but $13^3 = 2197 < 27720$.

If $m = 9$ or $10$, then $7 \cdot 5 \cdot 9 \cdot 8 = 2520 \mid n$, but $11^3 = 1331 < 2520$.

If $m = 8$, then $7 \cdot 5 \cdot 3 \cdot 8 = 840 \mid n$, but $9^3 = 729 < 840$.

If $m = 7$, then $7 \cdot 5 \cdot 3 \cdot 4 = 420 \mid n$ and $n < 8^3 = 512$. Therefore $n = 420$.