Let $k > \frac{m(n-1)}{n}$. Since $\frac{kn}{m} > n - 1$ at least one alternative obtained all $n$ votes. Therefore, any chosen $a$ obtained $n$ votes in profile $R$ and by definitions obtains also $n$ votes in profile $R'$. Thus, the condition $k > \frac{m(n-1)}{n}$ is sufficient for monotonicity.

Let $2 \le k \le \frac{m(n-1)}{n}$. We construct a profile $R$: for the first voter $a$ is the first, $b$ is the last. For the second, $b$ is the second, $a$ is the last. For all other voters, $a$ is the first, $b$ is the second. Each of $a$ and $b$ are not selected by exactly one voter. In our profile each alternative also is not selected by at least one voter. It is possible, since $n(m-k) \ge m$. No alternative obtained $n$ votes, $a$ and $b$ are chosen by getting $n-1$ votes.

The only difference between profiles $R'$ and $R$ is that the first voter charged the positions of the second alternative and $b$. Obviously, $R'$ $a$-dominates $R$, but $b$ obtained $n$ votes, $a$ obtained $n-1$ votes and is not selected. Thus, $k$-plurality choice rule is not monotonic.

Let $1 = k \le \frac{m(n-1)}{n}$. 1-plurality choice rule is monotonic if $n = 4, m = 3$: if $a$ is chosen, $a$ obtained at least 2 votes, in any other profile $a$ also gets at least 2 votes and other alternatives also get at most 2 votes! In other cases an example similar to the example above shows that the 1-plurality choice rule is not monotonic. Done.