How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$

If $n$ is divisible by $3$ there is nothing more to prove. Suppose that $n$ is not divisible by $3$ Then the remainder of dividing $n$ by $3$ is either $1$ or $2$. In the first case $n + 2$ is divisible by $3$; in the second case $n + 4$ is divisible by $3$.

$n+4$ is divisible by 3 if and only if $n+1$ is, so it's enough to consider the three consecutive numbers $n$, $n+1$ and $n+2$, one of which is divisible by 3.

$$ \begin{array}{c|lcr} n & \ n & \ n+2 & \ n+4 \\ \hline 3k & \fbox{$3k$} & 3k+2 & 3(k+1)+1 \\ 3k+1 & 3k+1 & \fbox{$3(k+1)$} & 3(k+1)+2 \\ 3k+2 & 3k+2 & 3(k+1)+1 & \fbox{$3(k+2)$} \\ \end{array} $$