Answer: 59

Solution:

If $k^3 \le n \lt (k+1)^3,$ then the interval $(n, mn]$ contains the cube $(k+1)^3$ as long as $(k+1)^3 \le m k^3,$ i.e. $\left(1 + \frac{1}{k}\right)^3 \le m.$ Since $\left(1 + \frac{1}{k}\right)^3 \le 8$ for all $k \ge 1,$ every $m \ge 8$ has $Q(m) = 1.$

For $4 \le m \le 7:$ $n = 1$ fails since $(1, m]$ contains no cube, but for $n \ge 2$ the interval works: $8 \le 4n$ covers $2 \le n \le 7,$ and $\left(1 + \frac{1}{k}\right)^3 \le \frac{27}{8} \lt 4$ covers $k \ge 2.$ So $Q(4) = Q(5)$ $= Q(6) = Q(7) = 2.$ For $m = 3:$ $n = 8$ fails (no cube in $(8, 24]$), while $27 \le 3n$ covers $9 \le n \le 26$ and $\left(\frac{4}{3}\right)^3 \lt 3$ covers $k \ge 3,$ so $Q(3) = 9.$ For $m = 2:$ $n = 31$ fails (no cube in $(31, 62]$), while $64 \le 2n$ covers $32 \le n \le 63$ and $\left(\frac{5}{4}\right)^3 \lt 2$ covers $k \ge 4,$ so $Q(2) = 32.$

Therefore $$\begin{aligned} &\sum_{m=2}^{2017} Q(m) \\ &\quad = 32 + 9 + 4 \cdot 2 + 2010 \cdot 1 \\ &\quad = 2059, \end{aligned}$$ and the remainder is $59.$