I feel like this proof should be easy.
Prove that for 3 consecutive natural numbers, 1 of the numbers will be divisible by 3.
Intuitively this makes sense (add 1, 2 or 3 and then it's divisible by 3, obviously).
let $\displaystyle n \in \mathbb{N}, A = \{n, n+1, n+2\} $
if $\displaystyle n \equiv 0 \mod{3} $, then $\displaystyle A \equiv \{0,1,2\} \mod{3} $
if $\displaystyle n \equiv 1 \mod{3} $, then $\displaystyle A \equiv \{1,2,0\} \mod{3} $
if $\displaystyle n \equiv 2 \mod{3} $, then $\displaystyle A \equiv \{2,0,1\} \mod{3} $
So in all three cases, $\displaystyle \exists \; a \in A $ s.t. $\displaystyle a \equiv 0 \mod{3} \Rightarrow a $ is divisible by three.