Answer: $\mathrm{C}$.
Solution: According to the problem, we have
$$
\begin{array}{c}
m=2017 \times\left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times \cdots \times\left(1-\frac{1}{1000}\right) \\
=2017 \times \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{999}{1000}=\frac{2017}{1000} \\
2<m \leqslant 3 .
\end{array}
$$

Obviously,
Therefore, the correct choice is C.
19. 2017 minus its $\frac{1}{2}$, then minus the remaining $\frac{1}{3}$, then minus the remaining $\frac{1}{4}$, and so on, until finally minus the remaining $\frac{1}{1000}$, the final result is $m$, then the range of $m$ is ( ).
(A). $03$