Proof For any integer $m \ge 2$, if the remainder of $x$ divided by $m$ is $i$, $i \in \{0, 1, \dots, m-1\}$, then $x$ belongs to the residue class modulus $m$, $K_i$.
Suppose in the set $A = \{a_1, a_2, \dots, a_n\}$, the number of elements that belong to $K_i$ is $n_i$ ($i = 0, 1, 2, \dots, m-1$), while in the set $B = \{1, 2, \dots, n\}$, the number of elements that belong to $K_i$ is $n'_i$ ($i = 0, 1, 2, \dots, m-1$), then
$$\sum_{i=0}^{m-1} n_i = \sum_{i=0}^{m-1} n'_i = n. \qquad \textcircled{1}$$
It is obvious that for every $i, j$, $|n_i' - n_j'| \le 1$, and $x-y$ is a multiple of $m$ if and only if $x, y$ belong to the same residue class. As to any two elements $a_i, a_j$ in $K_i$, we have $m \mid a_j - a_i$.
Hence, the $n_i$ elements in $K_i$ form $\binom{n_i}{2}$ multiples of $m$.
Considering all the $i$, we obtain
$$\bar{A}(m) = \sum_{i=0}^{m-1} \binom{n_i}{2}.$$
Similarly,
$$\bar{B}(m) = \sum_{i=0}^{m-1} \binom{n_i'}{2}.$$
Hence, to solve the problem, we just need to prove that
$$\sum_{i=0}^{m-1} \binom{n_i}{2} \ge \sum_{i=0}^{m-1} \binom{n_i'}{2}, \quad \text{and it can be simplified to}$$
$$\sum_{i=0}^{m-1} n_i^2 \ge \sum_{i=0}^{m-1} n_i'^2. \qquad \textcircled{2}$$
From ①, if for every $i, j$, $|n_i - n_j| \le 1$, then $n_0, n_1, \dots, n_{m-1}$ and $n'_0, n'_1, \dots, n'_{m-1}$ must be the same group (in spite of the different order), and equality holds in ②. Otherwise, if there exist $i, j$, such that $n_i - n_j \ge 2$, then we should just change the two elements $n_i, n_j$ for $\bar{n}_i, \bar{n}_j$ respectively, where $\bar{n}_i = n_i - 1$, $\bar{n}_j = n_j + 1$, and $n_i + n_j = \bar{n}_i + \bar{n}_j$. Since
$$(n_i^2 + n_j^2) - (\bar{n}_i^2 + \bar{n}_j^2) = 2(n_i - n_j - 1) > 0,$$
the sum of the left side in ② will decrease after adjustment. Therefore, the minimum value of ② is attained if and only if $n_0, n_1, \cdots, n_{m-1}$ and $n'_0, n'_1, \cdots, n'_{m-1}$ are the same group (in spite of the different order), that is, the inequality ② holds.