The numbers are $1,2, \cdots, 10$; there are also a big and a small joker, each numbered 0. Several cards are drawn from this deck, and the score is calculated according to the following rule: each card numbered $k$ is counted as $2^{k}$ points. If their total score is 2004, these cards are called a "good" card group. Try to find the number of "good" card groups.
For $n \in\{1,2, \cdots, 2004\}$, let $a_{n}$ denote the number of card sets with a total score of $n$, then $a_{n}$ is equal to the coefficient of $x^{n}$ in the expansion of the function
$$
f(x)=\left(1+x^{2^{0}}\right)^{2}\left(1+x^{2^{1}}\right)^{3} \cdots\left(1+x^{2^{10}}\right)^{3}
$$

(assuming $|x|<1$). Since
$$
\begin{aligned}
f(x) & =\frac{1}{1+x}\left\{\left(1+x^{2^{0}}\right)\left(1+x^{2^{1}}\right)\left(1+x^{2^{2}}\right) \cdots\left(1+x^{2^{10}}\right)\right\}^{3} \\
& =\frac{1}{(1+x)(1-x)^{3}}\left(1-x^{2^{11}}\right)^{3}=\frac{1}{\left(1-x^{2}\right)(1-x)^{2}}\left(1-x^{2^{11}}\right)^{3},
\end{aligned}
$$

and $n \leqslant 2004<2^{11}$, so $a_{n}$ is equal to the coefficient of $x^{n}$ in the expansion of $\frac{1}{\left(1-x^{2}\right)(1-x)^{2}}$. Since
$$
\begin{array}{l}
\frac{1}{\left(1-x^{2}\right)(1-x)^{2}}=\frac{1}{1-x^{2}} \cdot \frac{1}{(1-x)^{2}}-\left(1+x^{2}+x^{4}+\cdots+x^{2 k}+\cdots\right)(1+ \\
\left.2 x+3 x^{2}+\cdots+(2 k+1) x^{2 k}+\cdots\right),
\end{array}
$$

it follows that the coefficient of $x^{2 k}$ is
$$
a_{2 k}=1+3+5+\cdots+(2 k+1)=(k+1)^{2}, k=1,2, \cdots .
$$

Therefore, the number of "good" card sets is
$$
a_{2004}=1003^{2}=1006009 .
$$