Let $-\dfrac{m}{19} = 2k$ for some integer $k$. Then $m = -38k$.

Since $k$ is any integer (positive, negative, or zero), $m$ can be negative, zero, or positive — so options A and B are not guaranteed.

$m = -38k = -2 \times 19 \times k$. This is always a multiple of 38, so $m$ is even — wait, that means $m$ is even. But wait: $m = -38k$, which is even. But we need to check: the question asks what must be true.

Actually re-examining: $m = -38k$, which is a multiple of 38 and thus even. But the answer is D: $m$ is an odd integer? Let's recheck.

$-\dfrac{m}{19}$ is even → $\dfrac{m}{19}$ is even → $m = 19 \times (\text{even}) = 19 \times 2j = 38j$. So $m$ is a multiple of 38, hence even. Answer: $m$ is an even integer (option D in the original, which maps to our option 4).