We claim that if you have cards numbered $1,2, \ldots, 3^{2 n}$ and perform $2 n$ successive grouping operations, then $c$ is a possible value for your last remaining card if and only if $$3^{n} \leq c \leq 3^{2 n}-3^{n}+1$$ This gives $3^{2 n}-2 \cdot 3^{n}+2$ possible values of $c$, for a final answer of $3^{2004}-2 \cdot 3^{1002}+2$. Indeed, notice that the last remaining card $c$ must have been the largest of some set of three at the $(2 n-1)$ th step; each of these was in turn the largest of some set of three (and so $c$ was the largest of some set of 9 cards) remaining at the $(2 n-3)$ th step; each of these was in turn the largest of some set of three (and so $c$ was the largest of some set of 27 ) remaining at the $(2 n-5)$ th step; continuing in this manner, we get that $c$ was the largest of some $3^{n}$ cards at the first step, so $c \geq 3^{n}$. A similar analysis of all of the steps in which we save the smallest card gives that $c$ is the smallest of some set of $3^{n}$ initial cards, so $c \leq 3^{2 n}-3^{n}+1$. To see that any $c$ in this interval is indeed possible, we will carry out the groupings inductively so that, after $2 i$ steps, the following condition is satisfied: if the numbers remaining are $a_{1}<a_{2}<\cdots<a_{3^{2(n-i)}}$, then $c$ is one of these, and there are at least $3^{n-i}-1$ numbers smaller than $c$ and at least $3^{n-i}-1$ numbers larger than $c$. This is certainly true when $i=0$, so it suffices to show that if it holds for some $i<n$, we can perform the grouping so that the condition will still hold for $i+1$. Well, we first group the smallest numbers as $\left\{a_{1}, a_{2}, a_{3}\right\},\left\{a_{4}, a_{5}, a_{6}\right\}, \ldots,\left\{a_{3^{n-i}-5}, a_{3^{n-i}-4}, a_{3^{n-i}-3}\right\}$. We then group the remaining numbers in such a way that $c$ and the largest $3^{n-i}-1$ numbers are each the largest in its respective group; it is easy to see that we can do this. After retaining the largest number in each group, we will then have at least $3^{n-i-1}-1$ numbers smaller than $c$ remaining and at least $3^{n-i}-1$ numbers larger than $c$ remaining. And for the next grouping, we similarly group the largest $3^{n-i}-3$ numbers into $3^{n-i-1}-1$ groups, and arrange the remaining numbers so that the smallest $3^{n-i-1}-1$ numbers and $c$ are all the smallest in their groups. After this round of discarding, then $c$ will be retained, and we will still have at least $3^{n-i-1}-1$ numbers larger than $c$ and $3^{n-i-1}$ numbers smaller than $c$. This proves the induction step, and now the solution is complete.