Two people $A$ and $B$ play the following game: They take from $\{0, 1, 2, 3,..., 1024\}$ alternately  $512$, $256$, $128$, $64$, $32$, $16$, $8$, $4$, $2$, $1$, numbers away where $A$ first removes $512$ numbers, $B$ removes $256$ numbers etc. Two numbers $a, b$ remain ($a < b$). $B$ pays $A$ the amount $b - a$. $A$ would like to win as much as possible, $B$ would like to lose as little as possible. What profit does $A$ make if does every player play optimally according to their goals? The result must be justified.
To solve this problem, we need to analyze the game strategy for both players, $A$ and $B$, and determine the optimal moves they can make to maximize or minimize the profit, respectively. We will generalize the problem to a set $\{0, 1, \cdots, 4^n\}$ and show that the answer is $2^n$.

1. **Initial Setup and Generalization:**
   - The game starts with the set $\{0, 1, 2, \ldots, 1024\}$.
   - $A$ removes $512$ numbers, $B$ removes $256$ numbers, and so on.
   - We generalize the problem to a set $\{0, 1, \cdots, 4^n\}$.

2. **Strategy for Player $A$:**
   - $A$ aims to maximize the difference $b - a$.
   - Each turn, $A$ can remove the even indices in the sorted set.
   - By removing the even indices, $A$ ensures that the remaining numbers are more spread out, increasing the potential difference between the smallest and largest remaining numbers.
   - After $n$ turns, $A$ can increase the cost by at least $2^n$.

3. **Strategy for Player $B$:**
   - $B$ aims to minimize the difference $b - a$.
   - At the $i$-th turn of $B$, the size of the remaining set is $2^{2n-i} + 1$.
   - $B$ removes $2^{2n-i-1}$ elements, either the smallest or the largest elements.
   - By removing the smallest or largest elements, $B$ can ensure that the maximum difference in the set is reduced by at most half.
   - After $n$ turns, $B$ can decrease the cost by at most $2^n$.

4. **Combining the Strategies:**
   - Both players play optimally according to their goals.
   - $A$ can ensure that the difference $b - a$ is at least $2^n$.
   - $B$ can ensure that the difference $b - a$ is at most $2^n$.
   - Therefore, the optimal profit for $A$ is exactly $2^n$.

The final answer is $\boxed{2^n}$