If $A(n)$ is true, then since $2mn - 1 \ge 2n - 1$, among those integers there's $n$ of them which sum to a multiple of $n$, say $k_1n$, with $k_1 \in \mathbb{Z}$. Remove this group of $n$ integers to its own group, so you're left with $(2m - 1)n - 1$ integers. If $m \gt 1$, then $(2m - 1)n - 1 \ge 2n - 1$, so you can find another group of $n$ integers which sum to a multiple of $n$, say $k_2n$, with $k_2 \in \mathbb{Z}$. Once again, remove this group of $n$ integers into a separate group. You can repeat this $2m - 1$ times to get that same number of groups set aside, with the $i$'th (for each $1 \le i \le 2m - 1$) group having a sum of $k_i n$, for some $k_i \in \mathbb{Z}$.
Consider the $k_i$, for $1 \le i \le 2m - 1$, as a group of integers. If $A(m)$ is true, there's $m$ of the $k_i$ intgers which sum to a multiple of $m$, say $qm$, with $q \in \mathbb{Z}$. Take the $m$ groups of the associated $n$ integers, say with indices $i_j$ for $1 \le j \le m$, for a total of $mn$ integers, to get their total sum is
$$S = \sum_{j=1}^{2m-1}k_{i_j} n = n\sum_{j=1}^{2m-1}k_{i_j} = n(qm) = q(mn) \tag{1}$$
Thus, if $A(m)$ and $A(n)$ are true, you have $mn$ integers from among $2mn-1$ integers which sum to a multiple of $mn$, i.e., $A(mn)$ is also true.
FYI, this is an example of the Zero sum problem. Also, note the proof above is very similar to what is in the second page of the original paper of Erdős, Ginzburg and Ziv.
