Prove that one of any three consecutive positive integers must be divisible by $3$.

(N/A) Any three consecutive positive integers can be represented in the form $n, (n+1),$ and $(n+2)$,where $n$ is any positive integer $(n \in \mathbb{N})$.
By the Division Algorithm,any integer $n$ can be expressed in one of the forms $3q, 3q+1,$ or $3q+2$,where $q$ is a non-negative integer.
Case $1$: If $n = 3q$,then $n$ is divisible by $3$.
Case $2$: If $n = 3q + 1$,then $n + 2 = (3q + 1) + 2 = 3q + 3 = 3(q + 1)$,which is divisible by $3$.
Case $3$: If $n = 3q + 2$,then $n + 1 = (3q + 2) + 1 = 3q + 3 = 3(q + 1)$,which is divisible by $3$.
In all possible cases,one of the three consecutive integers $n, n+1,$ or $n+2$ is divisible by $3$.