This number has to have 2005 digits equal to 1. Notice that $\displaystyle 2005 = 333\times6+7$. So the number consists of 333 batches consisting of the digits 111111, together with one remaining batch that consists of seven 1s with a 7 somewhere between the 1s.
Next, notice that 111111 is divisible by both 7 and 13. Also, 11171111 is divisible by 7, and 11111711 is divisible by 13. So if the exceptional batch (the one containing the 7) has the 7 either between the third and fourth 1, or between the fifth and sixth 1, then the entire 2006-digit number will be composite.
So there are (at least) two composite numbers for each possible placing of the exceptional batch. But there are 334 such placings: the exceptional batch could come at the beginning or the end of the whole number, or between any consecutive pair of 111111s. That seems to give a total of at least 668 composite numbers, two more than was asked for.