Example 1 A deck of tri-color cards, consisting of 32 cards in total, where there are 10 cards of each color: red, yellow, and blue, numbered $1,2, \cdots, 10$; in addition, there is a big joker and a small joker, each numbered 0. From this deck, several cards are drawn, and the score is calculated according to the following rules: each card numbered $k$ is worth $2^{k}$ points. If the total score of these cards is 2004, they are called a "good card group." Try to find the number of good card groups.
【Analysis 1 calls this problem the "Two Kings Problem". If an additional wild card (called the "Middle King") is added, also numbered 0, and each is given a red, yellow, or blue color, referred to as the "Red King", "Yellow King", and "Blue King", it is then called the "Three Kings Problem".

First, in the Three Kings Problem, the number of card groups with a score of $n$ is denoted as $u_{n}$, and the scores of the red, yellow, and blue cards in each group are denoted as $x$, $y$, and $z$ respectively. Then, $x+y+z=n$. It is easy to see that the number of non-negative integer solutions to equation (1) is $\mathrm{C}_{n+2}^{2}$. For any set of determined $x$, $y$, and $z$, their binary representation is unique, thus, each non-negative integer solution to equation (1) corresponds to a unique card group. $\qquad$ Next, consider the Two Kings Problem. For $n \in\{1,2, 2004\}$, let the number of card groups with a score of $n$ be $a_{n}$. Find $a_{2}$.

For any card group with a score of $2k$, consider two cases:
(1) If the group contains no wild cards, then it is the no-wild-card case in the Three Kings Problem. In this case, if the score of each card is allowed to have wild cards, and each score is divided by 2, it corresponds to a card group in the Three Kings Problem. This correspondence is one-to-one, so the number of such card groups is $u_{k}$.
(2) If the group contains wild cards, since the sum of the scores is even, there must be two wild cards. Removing these two wild cards reduces the problem to the no-wild-card case in the Three Kings Problem with a score sum of $2k-2$. If the score of each card is divided by 2, it corresponds to a card group in the Three Kings Problem with a score of $k-1$, and this correspondence is one-to-one. Thus, the number of such card groups is $u_{k-1}$.
Therefore, $a_{2k}=u_{k}+u_{k-1}$
$=\frac{(k+1)(k+2)}{2}+\frac{k(k+1)}{2}=(k+1)^{2}$
In particular, when $2k=2004$, i.e., $k=1002$,
$a_{2004}=1003^{2}=1006009$.
【Analysis 2】Since $2004<2^{11}$, adding some cards with values $2^{11}, 2^{12}, \cdots$ does not affect the essence of the problem.
(1) Consider the Three Kings Problem.
For $n \in\{1,2, \cdots, 2004\}$, the number of ways to select several numbers from
the red set $A=$
the yellow set $B=\left\{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\right\}$

such that their sum is $n$ is denoted as $u_{n}$. Then
$u_{1}=3$ (since $1=1_{A}=1_{B}=1_{C}$)
$u_{2}=6$ (since $2=2_{A}=2_{B}=2_{C}=1_{A}+1_{B}=$
$+1_{C}=1_{C}+1_{A}$) Similarly, $u_{3}$
Similarly, $u_{3}=10, u_{4}=15, u_{5}=21, \cdots$
Define $u_{0}=1$
Generally, $u_{2n}=u_{n}+3u_{n-1}, u_{2n+1}=3u_{n}+u_{n-1}, \quad$ (3)
This is because
(i) For any card group with a total score of $2n$
If the group contains no wild cards, dividing each score by 2 results in a situation equivalent to a total score of $n$ and allowing wild cards. This type of card group has $u_{n}$ groups.
If the group contains wild cards, then there must be two wild cards. Removing these wild cards reduces the problem to a total score of $2n-2$ and no wild cards. Dividing each score by 2, this is equivalent to a total score of $n-1$ and allowing wild cards. Thus, this type of card group has $3u_{n-1}$ groups.
Therefore, $u_{2n}=u_{n}+3u_{n-1}$
(ii) For any card group with a total score of $2n+1$, it must contain an odd number of wild cards.
If the group contains one wild card, there are three choices. Removing the wild card reduces the problem to a total score of $2n$ and no wild cards. Dividing each score by 2, this is equivalent to a total score of $n$ and allowing wild cards. Thus, this type of card group has $3u_{n}$ groups.
If the group contains three wild cards, removing the wild cards reduces the problem to a total score of $2n-2$ and no wild cards. Dividing each score by 2, this is equivalent to a total score of $n-1$ and allowing wild cards. Thus, this type of card group has $u_{n-1}$ groups.
Therefore, $u_{2n+1}=3u_{n}+u_{n-1}$, using induction, from conclusion (3) we get
$u_{n}=\frac{(n+1)(n+2)}{2}$.
In fact, equation (4) has been verified for $n<4$.
Assume equation (4) holds for $n<2k$. Then
$u_{2k}=u_{k}+3u_{k-1}$
$=\frac{(k+1)(k+2)}{2}+3 \cdot \frac{k(k+1)}{2}$
$=\frac{(2k+1)(2k+2)}{2}$
$u_{2k+1}=3u_{k}+u_{k-1}$
$=3 \cdot \frac{(k+1)(k+2)}{2}+\frac{k(k+1)}{2}$
$=\frac{(2k+2)(2k+3)}{2}$
Thus, equation (4) also holds for $n<2k+2$.
Therefore, equation (4) holds for each $n$.
(2) Consider the Two Kings Problem (i.e., the problem in this question with only two wild cards).

For $n \in\{1,2, \cdots, 2004\}$, let $a_{n}$ be the number of card groups with a total score of $n$.

When $2k \leqslant 2004$, for any card group with a total score of $2k$:

If the group contains no wild cards, it is equivalent to the no-wild-card case in the Three Kings Problem. Dividing each score by 2 results in a situation equivalent to a total score of $k$ and allowing wild cards. This type of card group has $u_{k}$ groups.
If the group contains wild cards, it must be two wild cards (both the big and small wild cards are in the group). Removing the wild cards reduces the problem to the no-wild-card case in the Three Kings Problem with a total score of $2k-2$. Dividing each score by 2 results in a situation equivalent to a total score of $k-1$ and allowing wild cards. According to the above discussion, this type of card group has $u_{k-1}$ groups.
Thus, $a_{2k}=u_{k}+u_{k-1}$
$=\frac{(k+1)(k+2)}{2}+\frac{k(k+1)}{2}$
$=(k+1)^{2}$
where $k=1,2$,
Therefore, $a_{2004}=1003^{2}=1006009$, which is the answer.