= Solution
This is equality of remainders in <modular arithmetic>. It is reflexive because $3\mid(n-n)=0$, symmetric because $3\mid(n-m)$ implies $3\mid(m-n)$, and transitive because $3\mid(n-m)$ and $3\mid(k-n)$ imply $3\mid(k-m)$ by addition. Thus
$$
\boxed{m\sim n\iff m\equiv n\pmod3\text{ is an equivalence relation}.}
$$
Its three <equivalence classes> are
$$
\boxed{C_r=\{n\in\mathbb N:n\equiv r\pmod3\},\qquad r=0,1,2.}
$$
If zero is included in $\mathbb N$, it belongs to $C_0$; otherwise that class starts at three.
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