2. Let $n \geqslant 2$ be a positive integer, and let $A_{n}$ denote the set of all proper fractions with denominator $n$. Then, the number of elements in the set $A_{2008} \cap\left(\bigcup_{k=1}^{2007} A_{k}\right)$ is ( ).
(A) 2007
(B) 1008
(C) 1007
(D) 1000
2.C.

Since $2008=2^{3} \times 251$ (251 is a prime number), among the numbers $1,2, \cdots, 2007$, the numbers that are not coprime with 2008 include all the even numbers, as well as $251,3 \times 251,5 \times 251,7 \times$ 251. In total, there are $1003+4=1007$ such numbers. For each such number $a$, the proper fraction $\frac{a}{2008}$, when reduced to its simplest form, will appear in the set $\bigcup_{k=1}^{200} A_{k}$.
Therefore, the number of elements in $A_{2008} \cap\left(\bigcup_{k=1}^{2007} A_{k}\right)$ is 1007.