well i believe that the minimum should be 2008, it holds when we put four in a group and the minimum number of intersection points among 4 circles is 4. But... it's quite a hard problem to tackle (well i suppose it would not really be that hard coz CHKMO traditionally consist of moderate questions~~ probably trivial to most experienced mathlinkers~)

First by enumeration,it is easy to show that on each circl,there is at least three points.Now,if a point belongs to n circles ,then each of these circle get score$\frac{1}{n}$ from this point.Now, we will prove that the sum of score on each circle is no less than 1. let $A_{1}$,$A_{2}$,.......$A_{t}$ are all on the same circle,and belong to at least two circles so the circle passes $A_{i}$ is no more than t,so the score of this circle $\ge\frac{1}{t}*t$=1, Hence the score of all the circle is at least 2008,so there are at least 2008 points