Russia — Regional Round

Let $n$, $n+1$, $n+2$, $n+3$ be the given numbers. The sum of the three smallest of them is $3n + 3 = 3(n + 1)$, and the sum of the three largest numbers is $3(n + 2)$. But at least one of the numbers $n+1$ and $n+2$ is even, that is, equal to the product of $2$ and $k$, where $k > 3$. Therefore, this sum can be represented as the product of three distinct natural numbers: $2$, $3$, and $k$.