Not true. Just choose all of the points in the same place.

Yes, Your saying is true. I fix this. Thanks jmerry.

Suppose a point $A$ is at distance $r$ from the center $O$ of a circle of radius $1$. Choose a point $B$ on the circle at random. What is the average value of the distance $AB$? While this is easy to express as an integral, the integral in question is an elliptic integral that can't be calculated exactly except in a few special cases. Instead, we estimate. Consider two diametrically opposite points $X$ and $Y$ on the circle. By the triangle inequality, $XA+YA\ge XY=2$. Average over everything, dividing by $2$ since each point is covered twice, and the average of the distance $AB$ is at least $1$. (Equality comes when $A$ is on each diameter, which occurs only at the center.) Now, sum 100 different versions of this, for our 100 points. Average values add, and thus the average sum of distances to our points for a point chosen at random on the circle is at least $100$. This average must be matched or exceeded somewhere, and that's the point we seek.