Problem 787

For every natural number:

When divided by 3, the remainder is one of the values 0, 1, 2. Among these values, there are no four different ones. Therefore, among any four arbitrary natural numbers, there are two that leave the same remainder when divided by 3.

If $r$ is this remainder, then these two numbers are of the form $3 p + r$ and $3 q + r$ with natural numbers $p, q$. Their difference is thus $(3 p + r) - (3 q + r) = 3(p - q)$, which is divisible by 3.