Solution:
Given any infinite sequence of natural numbers, we can find a non-decreasing subsequence (proof below).

So suppose the three sequences are $a$, $b$, and $c$. Take a non-decreasing subsequence of $a_i$. Suppose it is $a_{i_1}$, $a_{i_2}$, $a_{i_3}$, ... . Now consider the infinite sequence $b_{i_1}$, $b_{i_2}$, ... . It must have a non-decreasing subsequence. Suppose it is $b_{j_1}$, $b_{j_2}$, ... . Now consider the infinite sequence $c_{j_1}$, $c_{j_2}$, ... . It must have a non-decreasing subsequence $c_{k_1}$, $c_{k_2}$, ... . Each of the three subsequences $a_{k_1}$, $a_{k_2}$, ... , $b_{k_1}$, $b_{k_2}$, ... , $c_{k_1}$, $c_{k_2}$, ... is non-decreasing. So we may take, for example, $m = k_2$ and $n = k_1$.

[Proof that any infinite sequence of natural numbers has a non-decreasing subsequence: if the original sequence is unbounded, then we can take a strictly increasing subsequence. If not, then since there are only finitely many possible numbers not exceeding the bound, at least one of them must occur infinitely often.]