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Let each student be Ai for i=1,2,...,1200 Randomly choose 22 of them, lets say A1, A2, A3..., A22, then if the intersection of these 22 set=K and lKl=1, for the remaining Ai, they must share this element as well, so contradiction So for any 22 of them, their union K has lKl>=2 Then choose 21 of them, and do the similar thing, you will get for any 21 of them, lKl>=3 and actually by MI, you will eventually have for any t of them, lKl>=24-t so we have k=lAil>=23 now if we denote each club be Ci, then for 24 students and 24 different clubs, the configuration below: A1={C2, C3,...,C24} A2={C1, C3,...,C24} ... A24={C1, C2,...,C23} clearly satisfies the above conditions and now if we set all the remaining students to be the same as any of them, WLOG, lets say A1, the new system satisfies the conditions as well. So minimum of k=23

$\definecolor{A}{RGB}{0,148,79}\color{A}\fbox{Lemma}$ For any sets $X,Y$ $$|X\cap Y|=|Y|\iff X\cap Y=Y\iff Y\subseteq X.$$$\definecolor{A}{RGB}{100,0,205}\color{A}\fbox{Solution}$ Let the students be numbered $1,2,...,1200$ and the set of clubs each joined named $A_1,A_2,...,A_{1200}$ respectively. We will prove inductively on $k$ the following: $\definecolor{A}{RGB}{255,0,205}\color{A}\star$ for all $\definecolor{A}{RGB}{255,0,205}\color{A}k\in\lbrace 1,2,...,23\rbrace$ and all different integers $\definecolor{A}{RGB}{255,0,205}\color{A}i_1,i_2,...,i_k\in \lbrace 1,2,...,1200\rbrace$ $$\definecolor{A}{RGB}{255,0,205}\color{A}|A_{i_1}\cap A_{i_2}\cap...\cap A_{i_k}|\ge 24 -k.$$For $k=23$ it's true by initial assumption. Assume there exists $k\in\lbrace 2,3,4,...,23\rbrace$ such that $\definecolor{A}{RGB}{255,0,205}\color{A}\star$ is true. We will prove it for $k-1$. $\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Inductive step}$ Assume contrary: there exist different integers $i_1,i_2,...,i_{k-1}\in \lbrace 1,2,...,1200\rbrace$ such that $$|A_{i_1}\cap A_{i_2}\cap...\cap A_{i_{k-1}}|\le 24 -k.$$Then for any $j\in \lbrace 1,2,...,1200\rbrace\setminus\lbrace i_1,i_2,...,i_{k-1}\rbrace$ we have by inductive assumption $$1\le 24-k\le |A_{i_1}\cap A_{i_2}\cap...\cap A_{i_{k-1}}\cap A_j|\le |A_{i_1}\cap A_{i_2}\cap...\cap A_{i_{k-1}}|\le 24 -k.$$This and lemma imply $$A_j\supseteq A_{i_1}\cap A_{i_2}\cap...\cap A_{i_{k-1}}\neq \emptyset.$$Remember this is for all $j\in \lbrace 1,2,...,1200\rbrace\setminus\lbrace i_1,i_2,...,i_{k-1}\rbrace$ so there exists a club which is a common element of every set among $A_1,A_2,...,A_{1200}$. This is contradiction with assumed $A_1\cap A_2\cap ...\cap A_{1200}=\emptyset.\blacksquare$ $\definecolor{A}{RGB}{255,105,0}\color{A}\fbox{Conclusion}$ $$\definecolor{A}{RGB}{255,105,0}\color{A}\forall_{i\in\lbrace 1,2,...,1200\rbrace}|A_i|\ge 23.$$$\definecolor{A}{RGB}{255,0,0}\color{A}\fbox{Construction}$ Name the clubs $C_1,C_2,...,C_{24}$ and take $$\forall_{i\in\lbrace 1,2,...,24\rbrace}A_i=\lbrace C_1,C_2,...,C_{24}\rbrace\setminus \lbrace C_i\rbrace,\ \forall_{i\in\lbrace 24,25,...,1200\rbrace} A_i=A_1.$$#1778