China Mathematical Olympiad

Let $A=\{a_1,\dots ,a_m\}$ with $a_1>\cdots >a_m$. By (1) we have $a_1+\cdots +a_m=0$. Consider the smallest element of each $A_i$, the sum of these numbers is greater than $0$. Assume that there are exactly $k_i$ sets among $A_1,\dots ,A_n$ whose minimal element is $a_i$, $i=1,2,\dots ,m$. Then, one has $$k_1+\cdots +k_m=n.$$ By (2), we have $$k_1{a}_1+\cdots +k_m{a}_m>0.$$ For $s=1,2,\dots ,m-1$, there are in total $k_1+\cdots +k_m$ sets, whose minimal elements are greater than or equal to $a_s$. Therefore, the union of these sets is contained in $\{a_1,\dots ,a_m\}$, whence the number of elements does not exceed $s$.

Next, we prove that there exists $s\in \{1,2,\dots ,m-1\}$ such that $k=k_1+\cdots +k_s>\frac{sn}{m}$. We prove this claim by contradiction. Suppose that $$k_1+\cdots +k_s\le \frac{sn}{m},s=1,2,\dots ,m-1.$$ With the help of the Abel transform and the fact that $a_s-a_{s+1}>0$, $1\le s\le m-1$, we know that $$\begin{array}{rl}0<& \sum _{j=1}^m{k}_j{a}_j\\ =& \sum _{s=1}^{m-1}(a_s-a_{s+1})(k_1+\cdots +k_s)+a_m(k_1+\cdots +k_m)\\ \le & \sum _{s=1}^{m-1}(a_s-a_{s+1})\frac{sn}{m}+a_{m}n\\ =& \frac{n}{m}\sum _{j=1}^m{a}_j=0.\end{array}$$ We then get a contradiction. For such an $s$, we take the sets among $A_1,\dots ,A_n$, whose minimal elements are greater than $a_s$, say $A_{i_1},A_{i_2},\dots ,A_{i_k}$. Then, by the above results, we know that the total number of such sets is $k=k_1+\cdots +k_s>\frac{sn}{m}$, and the number of elements of their union does not exceed $s$, i.e., $$|A_{i_1}\cup A_{i_2}\cup \cdots \cup A_{i_k}|\le s<\frac{km}{n}=\frac{k}{n}|A|.$$ $\square$