On the table, there are $1024$ marbles and two students, $A$ and $B$, alternatively take a positive number of marble(s). The student $A$ goes first, $B$ goes after that and so on. On the first move, $A$ takes $k$ marbles with $1 < k < 1024$. On the moves after that, $A$ and $B$ are not allowed to take more than $k$ marbles or $0$ marbles. The student that takes the last marble(s) from the table wins. Find all values of $k$ the student $A$ should choose to make sure that there is a strategy for him to win the game.
1. **Initial Setup and Constraints**:
   - There are 1024 marbles on the table.
   - Two students, \(A\) and \(B\), take turns to take a positive number of marbles.
   - \(A\) goes first and takes \(k\) marbles where \(1 < k < 1024\).
   - On subsequent moves, neither \(A\) nor \(B\) can take more than \(k\) marbles.
   - The student who takes the last marble(s) wins.

2. **Winning Strategy Analysis**:
   - To ensure a winning strategy, \(A\) must leave a number of marbles that forces \(B\) into a losing position.
   - Consider the game modulo \(k+1\). If \(A\) can always leave a number of marbles that is congruent to 0 modulo \(k+1\), then \(A\) can force \(B\) into a losing position.

3. **Modulo Analysis**:
   - Initially, there are 1024 marbles. We need to analyze \(1024 \mod (k+1)\).
   - If \(1024 \equiv 0 \pmod{k+1}\), then \(A\) can take \(k\) marbles, leaving \(1024 - k\) marbles. After this, \(A\) can always play symmetrically to \(B\) modulo \(k+1\).

4. **Finding Suitable \(k\)**:
   - We need \(k+1\) to be a divisor of 1025 (since \(1024 + 1 = 1025\)).
   - The divisors of 1025 are 1, 5, 25, 41, and 1025.
   - Since \(1 < k < 1024\), we exclude 1 and 1025.
   - Therefore, the possible values for \(k+1\) are 5, 25, and 41, which correspond to \(k = 4, 24, 40\).

5. **Verification**:
   - For \(k = 4\), \(k+1 = 5\):
     \[
     1024 \equiv 4 \pmod{5}
     \]
     \(A\) can take 4 marbles, leaving 1020 marbles, which is \(0 \pmod{5}\). \(A\) can then play symmetrically to \(B\).
   - For \(k = 24\), \(k+1 = 25\):
     \[
     1024 \equiv -1 \pmod{25}
     \]
     \(A\) can take 24 marbles, leaving 1000 marbles, which is \(0 \pmod{25}\). \(A\) can then play symmetrically to \(B\).
   - For \(k = 40\), \(k+1 = 41\):
     \[
     1024 \equiv -1 \pmod{41}
     \]
     \(A\) can take 40 marbles, leaving 984 marbles, which is \(0 \pmod{41}\). \(A\) can then play symmetrically to \(B\).

Conclusion:
\[
\boxed{k = 4, 24, 40}
\]