3. [5] For which natural numbers $n$ can one find $n$ consecutive natural numbers whose product is equal to the sum of (possibly different) $n$ consecutive natural numbers?

Boris Frenkin
Answer: for odd $n$.

Solution. The product of $n$ consecutive integers is divisible by $n$, so the sum of these integers is also divisible by $n$. Therefore, the arithmetic mean of these numbers is an integer. This means that $n$ is odd. Here is an example for $n=2 m+1:((2 m)!-m)+((2 m)!-m+1)+\ldots+((2 m)!+m)=(2 m+1) \cdot(2 m)!=n!$ !