India — Team Selection Test, 2006

We use induction on $n$. The case $n = 1$ is immediate. Assume the result for $n$ and consider the case of $n+1$. Thus $A_1, A_2, \dots, A_{n+1}$ are subsets of $S$ with $|A_j| = 8$.
Now if we consider the sets $A_1, A_2, \dots, A_{j-1}, A_{j+1}, \dots, A_{n+1}$, these satisfy induction hypothesis for each $j = 1, 2, 3, \dots, n+1$. Hence there exists
$$x_j \in A_1 \cap A_2 \cap \dots \cap A_{j-1} \cap A_{j+1} \cap \dots \cap A_{n+1},$$
for each $j = 1, 2, \dots, n+1$. If $x_1, x_2, \dots, x_{n+1}$ are not all distinct, then we are through. Suppose they are all distinct. Consider $B = A_1 \cup A_2 \cup \dots \cup A_{n+1}$. We show that $|B| \le 25$. Note that each $x_j$ is over counted $n-1$ times in the union. Thus
$$|B| \le |A_1| + |A_2| + \dots + |A_{n+1}| - (n-1)(n+1) = 8(n+1) - (n^2-1) \le 25,$$
since $n^2 - 8n + 16 = (n-4)^2 \ge 0$. Thus the condition (i) fails. Clearly, the condition (2) also fails. Hence (iii) must hold.