Solution:

The answer is (D). Let us call $S(n)$ the sum of 2002 consecutive odd numbers starting from $n$.
$$S(1)=1+3+\ldots+4003=(1+4003)+(3+4001)+\ldots+(2001+2003)=4004 \cdot 1001 .$$
We note that $S(n+2)$ and $S(n)$ have 2001 addends in common and that their difference is therefore equal to
$$((n+2)+4002)-n=4004$$
The greatest common divisor among the $S(n)$ is therefore 4004.

SECOND SOLUTION.
The sum of the first $n$ odd numbers from 1 to $2 n-1$ is $n^{2}$, so the sum of 2002 consecutive odd numbers starting from $2 n+1$ is $(n+2002)^{2}-n^{2}=2002(2 n+2002)=4004(n+1001)$ so they are all multiples of 4004, and two consecutive ones have GCD 4004 because $\operatorname{GCD}(k, k+1)=1$.