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.

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

A somewhat silly answer:

You probably know that $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$; if not these can be proven easily via induction. Consider then \begin{align*} \sum_{i=1}^n (i^2 + 3i + 1) &= \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n \\ &= \frac{(2n^3 + 3n^2 + n) + 9(n^2+n) + 6n}{6} \\ &= \frac{2n^3 +12 n^2 + 16 n}{6} \\ &= \frac{n^3 + 6n^2 + 8n}{3} \\ &= \frac{n(n+2)(n+4)}{3} \end{align*}

On the left hand side, you have a sum of integers, so the right hand side is an integer. Hence $3 | n(n+2)(n+4)$, and since $3$ is prime it divides one of the three factors.