Example 3 A deck of tri-color cards, totaling 32 cards, includes 10 cards of each color: red, yellow, and blue, numbered $1,2, \cdots, 10$; there are also a big joker and a small joker, each numbered 0. Several cards are drawn from this deck, and their scores are calculated according to the following rule: each card numbered $k$ is worth $2^{k}$ points.

If the total score of these cards is 2004, then these cards are called a "good card group." Find the number of good card groups.
$(2004$, Girls' Mathematical Olympiad)
Solution: This problem is referred to as the "Two Kings Problem." If an additional joker (called the "Middle King") with a value of 0 is added, and each card is still assigned a value as described, it is called the "Three Kings Problem."

Since $2004 < 2^{11}$, adding some cards with values $2^{11}, 2^{12}, \cdots$ does not affect the essence of the problem.
First, consider the Three Kings Problem.
For $n \in \{1, 2, \cdots, 2004\}$, take some numbers from
the Red Set $A = \{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\}$,
the Yellow Set $B = \{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\}$,
the Blue Set $C = \{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\}$,

such that their sum is $n$. Let the number of ways to do this be denoted as $u_{n}$. Then
$$
\begin{array}{l}
u_{1} = 3 \left(1 = 1_{A} = 1_{B} = 1_{C}\right), \\
u_{2} = 6 \left(2 = 2_{A} = 2_{B} = 2_{C} = 1_{A} + 1_{B} = 1_{B} + 1_{C} = 1_{C} + 1_{A}\right).
\end{array}
$$

Similarly, $u_{3} = 10, u_{4} = 15, u_{5} = 21, \cdots \cdots$
Define $u_{0} = 1$. Generally, we have
$$
u_{2 n} = u_{n} + 3 u_{n-1}, \quad u_{2 n+1} = 3 u_{n} + u_{n-1}.
$$

This is because
(1) For a group of cards with a total score of $2 n$.
(i) If the group contains no jokers, dividing each value by 2 results in a situation equivalent to a total score of $n$ and allowing jokers. There are $u_{n}$ such groups.
(ii) If the group contains jokers, there must be two jokers. These two jokers can be paired in three ways. Removing the jokers, the situation is equivalent to a total score of $2 n - 2$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $n - 1$ and allowing jokers. There are $3 u_{n-1}$ such groups.
Therefore, $u_{2 n} = u_{n} + 3 u_{n-1}$.
(2) For a group of cards with a total score of $2 n + 1$, it must contain an odd number of jokers.
(i) If the group contains one joker, it has three choices.
Removing the joker, the situation is equivalent to a total score of $2 n$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $n$ and allowing jokers. There are $3 u_{n}$ such groups.
(ii) If the group contains three jokers, removing the jokers, the situation is equivalent to a total score of $2 n - 2$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $n - 1$ and allowing jokers. There are $u_{n-1}$ such groups.
Therefore, $u_{2 n+1} = 3 u_{n} + u_{n-1}$.
Using induction, from (1) we get
$$
u_{n} = \frac{(n+1)(n+2)}{2}.
$$

In fact, (2) has been verified for $n < 4$.
Assume that (2) holds for $n < 2 k$. Then
$$
\begin{array}{l}
u_{2 k} = u_{k} + 3 u_{k-1} \\
= \frac{(k+1)(k+2)}{2} + 3 \cdot \frac{k(k+1)}{2} \\
= \frac{(2 k+1)(2 k+2)}{2}, \\
u_{2 k+1} = 3 u_{k} + u_{k-1} \\
= 3 \cdot \frac{(k+1)(k+2)}{2} + \frac{k(k+1)}{2} \\
= \frac{(2 k+2)(2 k+3)}{2}.
\end{array}
$$

Thus, (2) also holds for $n < 2 k + 2$.
Therefore, (2) holds for every $n$.
Now consider the Two Kings Problem.
For $n \in \{1, 2, \cdots, 2004\}$, let $a_{n}$ be the number of card groups with a total score of $n$. When $2 k \leqslant 2004$, for any card group with a total score of $2 k$:
(1) If the group contains no jokers, it is equivalent to a situation in the Three Kings Problem with a total score of $2 k$ and no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $k$ and allowing jokers. There are $u_{k}$ such groups.
(2) If the group contains jokers, there must be two jokers (both the big and small jokers are in the group). Removing the jokers, the situation is equivalent to a total score of $2 k - 2$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $k - 1$ and allowing jokers. According to the above discussion, there are $u_{k-1}$ such groups.
Therefore, $a_{2 k} = u_{k} + u_{k-1}$
$= \frac{(k+1)(k+2)}{2} + \frac{k(k+1)}{2}$
$= (k+1)^{2} \quad (k=1, 2, \cdots)$.
Thus, $a_{2004} = 1003^{2} = 1006009$, which is the number of good card groups.