, 2023

Let $k$ be the number of seminars the club can organize. In the $i$-th seminar with $1 \le i \le k$, denote $A_i$, $B_i$, $C_i$ as the collection of participants presenting Math, Physics, Chemistry respectively. Then, for each $i$, we have $A_i$, $B_i$, $C_i$ are the partition of the members of the club. Replace $2023 \to n$.

In each seminar, the members will have $3$ ways to choose subjects to present, so there will be $3^k$ ways to choose in total. If $3^k < n$ then, according to the pigeonhole principle, there will be two members who must have the same choice in all $k$ seminars. However, the two members will always present the same subject for every seminar, contradiction. So there must be $n \le 3^k$.

We will show that for $n \le 3^k$ it is always possible to construct. Indeed, converting the members' numbers from $0, 1, 2, \ldots, n-1$ to base $3$, each number will correspond to a ternary string of length no more than $k$. To synchronize those strings, we add $0$ in front so that all strings have the same length $k$. We define the following rule: in the $i$th seminar with $1 \le i \le k$, all members having $0$ at position $i$ on their string will present Math, similarly, the same $1$ for Physics and the same $2$ for Chemistry.

Note that any two strings are different, so they must be different at a certain position, and the index of that position will be the index of the seminar in which the two respective members present $2$ different subjects, satisfying the constraint.

For $n = 2023$, we have $k \ge \lceil \log_3 2023 \rceil = 7$ and $k = 7$ is the minimum value. $\square$