Solution:

The $n$th term of $S_k$ is $1 + (n-1)k,$ so $2005$ is a term exactly when $(n-1)k = 2004$ for some positive integer $n,$ that is, exactly when $k$ divides $2004.$ Every divisor works, since $n - 1 = \frac{2004}{k}$ is then a positive integer.

Since $2004 = 2^2 \cdot 3 \cdot 167,$ the number of divisors is $(2+1)(1+1)(1+1) = 12.$