Suppose the points are placed on the $x$-axis with coordinates $(i, 0)$, $i = 0, \dots, 2016$. Then midpoints are $(i/2, 0)$, $i = 1, 2, \dots, 4031$. Thus there are $4031$ distinct midpoints.

Next we shall prove that there are at least $4031$ distinct midpoints. Let $A_1, \dots, A_{2017}$ be the points and assume that $A_1, A_2$ are the pair that are furthest apart. Consider the $4030$ segments from $A_1$ and $A_2$ to $A_3, \dots, A_{2015}$. The midpoints are distinct. For if $X, Y$ are two points so that the midpoints of $A_1X$ and $A_2Y$ coincide, then we have two cases. If the four points are not collinear, then they are vertices of a parallelogram with $A_1X, A_2Y$ as diagonals and $A_1A_2$ as a side. This is not possible as the longer diagonal is longer than a side. Otherwise $A_1, A_2, X, Y$ are collinear. Then it is easy to verify that if $X$ is in the segment $A_1A_2$, then $Y$ must be outside making $A_1A_2 < A_2Y$, a contradiction. Also none of these midpoints is the midpoint of $A_1A_2$. Thus we have at least $4031$ distinct midpoints.

In conclusion, the minimum number of midpoints is $4031$.