Denote two given points by zero and one on the real number line. Note that after we use the plotter $n$ times, we will obtain points with coordinate $\frac{a}{2^n}<1$ for some positive integer $a$ (this can be proved by induction, but it's obvious). So the distance between two points after $n^{th}$ use must also in the form $\frac{a}{2^n}$ for some positive integer $a$. We want to find positive integer $a,n$ such that $\frac{1}{2017}<\frac{a}{2^n}<\frac{1}{2016}$. And we also want to minimize $n$. Some calculations give us $a=65,n=17$. We need to show that $17$ times is enough, since $65=2^6+1$ Construct $\frac{1}{2},\frac{1}{4},...,\frac{1}{64},\frac{\frac{1}{64}+1}{2}=\frac{65}{2^7},\frac{65}{2^8},...,\frac{65}{2^{17}}$, done.

Solved here.