I can't understand, can you post the origin problem here?

maybe we can change "each of the 12 students forming the advisory board" into "one of the 12 students forming the advisory board "?

What do you mean by origin problem? It is in Turkish. No, I checked the Turkish version. Every one of 12 members of the board should be present so as to have a meeting.

So， you mean 12 arbitrary students could form an advisory board of a club，is that true？

Even the original of the problem is very confusing but I thought it was meant to say something like The students that are in any subset that has at least 12 students are able to make exactly 1 meeting.

Here is a proof that there must be $\boxed{\binom{2003}{11}}$ student clubs with $27$ student clubs. Let there be $x_i$ clubs with $i$ students in them, for each $i \ge 12.$ Let's count the number of sets of $x+12$ students, for $x \in [0, 2007]$, which can conduct an advisory meeting in two different ways. On one hand, it is $\binom{2019}{x+12}$ from the condition of the problem. On the other hand, each student club with $i$ students contributes $\binom{i-12}{x}$ to this. Hence, we obtain that: $$\sum_{i = 12}^{2019} x_i \cdot \binom{i-12}{x} = \binom{2019}{x+12}, \forall 0 \le x \le 2007.$$ With this result, we are ready to prove by induction that $x_i = \binom{2030-i}{11}$ for all $12 \le i \le 2019,$ from which our claimed result follows. We will prove this by downwards induction on $i$. The base case $i = 2019$ is immediate by plugging $x = 2007$ into the above. To complete the inductive step, it suffices to show that: $$\sum_{i = t+12}^{2019} \binom{i-12}{t} \cdot \binom{2030-i}{11} = \binom{2019}{t+12}, \forall 0 \le t \le 2007$$ We will do this by a nice combinatorial argument. Consider the lattice paths from $(0, 0)$ to $(2007-t, t+12).$ Divide them into cases based on the $y-$value of the first time that they cross the line $x = t + \frac12.$ Double-counting finishes from here. The induction is now complete, and so we've shown that $x_{27} = \binom{2030-27}{11} = \boxed{\binom{2003}{11}}.$ $\square$