Vietnamese Mathematical Olympiad

a) If $m=1$, then the class will have 8 group to attend, this is contradiction. If $m\ge 3$, consider any 3 groups $X_1,X_2,X_3$ then $$n\ge |X_1\cup X_2\cup X_3|\ge |X_1|+|X_2|+|X_3|-|X_1\cap X_2|-|X_2\cap X_3|-|X_1\cap X_3|\ge 4+4+4-1-1-1=9,\text{ (contradiction).}$$ So $m=2$, we can take a simple case that each group has 4 participating classes and no class joins 2 groups.

b) Denote by $S$ the number of sets $(A,\{B,C\})$ in which class $A$ has students participating in groups $B$ and $C$. There are $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ pairs $B,C$ and any two groups have no more than 4 co-participating class so we have $S\le 15\cdot 4=60$. Let $a_1,a_2,\dots ,a_n$ be the number of groups that the student of class 1, 2, ..., $n$ participate to. There are $6\cdot 10=60$ in total of participants so $a_1+a_2+\cdots +a_n=60$. We have $$\begin{array}{rl}S& =\left(\genfrac{}{}{0ex}{}{a_1}{2}\right)+\left(\genfrac{}{}{0ex}{}{a_2}{2}\right)+\left(\genfrac{}{}{0ex}{}{a_3}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{a_n}{2}\right)\\ & =\frac{a_1\cdot (a_1-1)+a_2\cdot (a_2-1)+\cdots +a_n\cdot (a_n-1)}{2}\\ & =\frac{1}{2}(a_1^2+a_2^2+\cdots +a_n^2)-30.\end{array}$$ Since $S\le 60$, we have $$a_1^2+a_2^2+\cdots +a_n^2\le 2(60+30)=180.$$ Applying the Cauchy-Schwarz inequality, we have $$n(a_1^2+a_2^2+\cdots +a_n^2)\ge (a_1+a_2+\cdots +a_n{)}^2=3600.$$ Hence $\frac{3600}{n}\le 180$, which implies that $n\ge 20$.

c) There are a total of $20\cdot 4=80$ participation so there will be a class $D$ with the number of student is at least $\lfloor \frac{80}{n}\rfloor$. And the groups that class $D$ participates in will all have the same 1 co-participant class (class $D$) and the remaining 3 classes of these groups are distinct. Therefore, we get $$n\ge 3\lfloor \frac{80}{n}\rfloor +1\ge 3\cdot \frac{80}{n}+1⟹n^2-n-240\ge 0.$$ This implies that $n\ge 16$. We can show a specific case with $n=16$ as following $$\begin{array}{rl}{A}_1& =\{1,2,3,4\},A_2=\{1,5,6,7\},A_3=\{1,8,9,10\},\\ A_4& =\{1,11,12,13\},A_5=\{1,14,15,16\},A_6=\{2,4,9,13\},\\ A_7& =\{2,10,12,14\},A_8=\{2,6,8,16\},A_9=\{2,7,11,15\},\\ A_{10}& =\{3,5,12,16\},A_{11}=\{3,8,13,15\},A_{12}=\{3,6,10,11\},\\ A_{13}& =\{3,7,9,14\},A_{14}=\{4,5,10,15\},A_{15}=\{4,7,8,12\},\\ A_{16}& =\{4,9,11,16\},A_{17}=\{4,6,13,14\},A_{18}=\{5,8,11,14\},\\ A_{19}& =\{6,9,12,15\},A_{20}=\{7,10,13,16\}.\end{array}$$ So in this case, the minimum value of $n$ is 16. $\square$