To prove that for any integer modulo $m$, it must be congruent to one of the following $m$ integers:
$$0,1, \cdots, m-1$$

Let $r_{i}$ (where $i=1,2, \cdots, m$) be an integer satisfying the condition
$$a_{i} \equiv r_{i}(\bmod m), \quad 0 \leqslant r_{i} \leqslant m-1,$$

Then we have
$$a_{1} \equiv r_{1}(\bmod m), a_{2} \equiv r_{2}(\bmod m), \cdots, a_{m} \equiv r_{m}(\bmod m) .$$
where $0 \leqslant r_{1} \leqslant m-1,0 \leqslant r_{2} \leqslant m-1, \cdots, 0 \leqslant r_{m} \leqslant m-1$. Since (3) and the assumption that any two integers chosen from $a_{1}, a_{2}, \cdots, a_{m}$ are not congruent modulo $m$, it follows that any two integers chosen from $r_{1}, r_{2}, \cdots, r_{m}$ are also not congruent modulo $m$. Therefore, $r_{1}, r_{2}, \cdots, r_{m}$ and $0,1, \cdots, m-1$ differ only in order, meaning that $a_{1}, a_{2}, \cdots, a_{m}$ form a complete residue system modulo $m$.