Solution:
(a) The answer is yes. The following picture shows that there is a nice set consisting of exactly 12 circles. It is also possible to construct a nice set with 22 circles (the picture can be found below). Since $2006 = 158 \cdot 12 + 5 \cdot 22$ we can make a nice set of 2006 by making a union of 158 disjoint nice sets of 12 circles each, and 5 disjoint nice sets each of which contains 22 circles.

(b) We will prove that a nice set can't contain an odd number of circles. Let $n$ be the total number of circles. We will count the number of pairs $(k, P)$ where $k$ is a circle in the set and $P$ a point at which $k$ touches another circle. For each circle $k$ there are exactly 5 such pairs, and hence the total number of pairs is $5n$. For each point $P$ there are exactly 2 pairs corresponding to it. Hence the total number of pairs has to be even, but $5n$ can't be even if $n$ is odd.