The answer is $k = 23$.

We first show $k \le 23$. We list the students as $S_1, S_2, \dots, S_{1200}$ and the clubs as $C_1, C_2, \dots, C_{24}$. Consider the following construction. For $1 \le j \le 24$, student $S_j$ joins clubs $C_1, \dots, C_{j-1}, C_{j+1}, \dots, C_{24}$. For $25 \le j \le 1200$, student $S_j$ joins the same clubs as $S_1$. Then we can see every $23$ students have a common club, but all $1200$ students do not have a common club.

Now we show $k \ge 23$. Suppose a student $S$ joins clubs $C_1, C_2, \dots, C_k$. Because all $1200$ students do not join a common club, for each $1 \le j \le k$, there is a student $S_j$ not joining the club $C_j$. Hence the students $S, S_1, S_2, \dots, S_k$ do not have a common club, forcing $k \ge 23$. The proof is complete.