After the first move, there are $1024-k$ marbles remaining and both students $A, B$ cannot take more than $k$ marbles during their turn.

We shall prove that if $k+1 \mid 1024-k$ then the student $A$ will have the strategy to win the game.

Indeed, let $1024-k = m(k+1)$ with $m \in \mathbb{Z}^{+}$. On each turn, if $B$ chooses $x$ marble(s) then $A$ will choose $k+1-x$ marble(s) on his next turn. Since $x \in \{1,2,3, \ldots, k\}$ then $k+1-x \in \{1,2,3, \ldots, k\}$ which implies that $A$ always can perform his turn. After $m$ turns of $A$ and $m$ turns of $B$, student $A$ takes the last marble then wins the game.

If $1024-k$ is not divisible by $k+1$ then we put $1024-k = m(k+1) + n$ with $0 < n < k+1$. In this case, the student $B$ can take $n$ marbles on the next turn. The number of remaining marbles is $m(k+1)$ then by applying the same argument above, the student $B$ has the strategy to win the game.

Hence, the condition we have to find is $k+1 \mid 1024-k$ or $k+1 \mid 1025$. Note that the divisors of $1025$ are $1,5,25,41,205,1025$ then
$$k+1 \in \{1,5,25,41,205,1025\} \text{ and } 0<k<1024.$$
This means that all values of $k$ we need to find are $4,24,40,204$.

$\square$