( this solution is a try) If a circle touches at least three sides internally, it is the inscribed circle (because there will only one point the intersection of two bissectrices and hence being $I$) If a circle touch a side externally, it won't touch an other of the reglular polygon and there will be maximum $2$ other sides who are tangent (when we take the line longer). In the other cases, it will touch maximum $2$ sides. We can have maximum have one inscribed circle (if we take it twice, each tangent satisfy and the answer would be a stupid $2$). Because it has to be tangent to $2$ circles, the next circles have to touch together again $2011$ sides, so we need at least $671$ and that goes, so the answer is $672.$