Assume there are only the distances $D=\{d_1,d_2,\hdots,d_{31}\}$ and consider two of the points $A,B$. Also, draw the $62$ circles centered at $A,B$ with radii $d_i$. Since each other point in the set has a distance to $A$ and $B$ which is in $D$, it has to be the intersection of two such circles. However, the two circles intersect in at most $2\cdot 31^2=1922$ points.

What about when the points are 2016 and you have to prove that there must be at least 45 different values? It's from a different source (but i don't remember which one it is), and in this case the above proof seems not to work.