Problem 906

3. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, a_{1}>a_{2}>\cdots>a_{m}$. Then, by condition (1), we have
$$a_{1}+a_{2}+\cdots+a_{m}=0 .$$

Consider the smallest number in each $A_{i}$, and let $A_{1}, A_{2}$, $\cdots, A_{n}$ contain exactly $k_{i}$ sets whose smallest number is $a_{i}(i=1$, $2, \cdots, m$ ).
Thus, $\sum_{i=1}^{m} k_{i}=n$, and by condition (2) we have
$$\sum_{j=1}^{m} k_{j} a_{j}>0 \text {. }$$

For $s(1 \leqslant s \leqslant m-1)$, there are $\sum_{i=1}^{3} k_{i}$ sets, all of whose smallest numbers are greater than or equal to $a_{x}$. Therefore, the union of these sets is contained in $\left\{a_{1}, \cdots, a_{3}\right\}$, and the number of elements is at most $s$.
Next, we prove by contradiction:
There exists $s(1 \leqslant s \leqslant m-1)$, such that
$$k=\sum_{i=1}^{s} k_{i}>\frac{s n}{m} .$$

Assume that for $s(1 \leqslant s \leqslant m-1)$, we have
$$\sum_{i=1}^{3} k_{i} \leqslant \frac{s n}{m} \text {. }$$

By Abel's transformation (note that $a_{1}-a_{1+1}>0$, $1 \leqslant s \leqslant m-1$ )
\begin{aligned} 0 & \frac{s n}{m}

There are $k$ sets, and the union of these sets has at most $s$ elements, i.e.,
$$\left|A_{i_{1}} \cup A_{i_{2}} \cup \cdots \cup A_{i_{k}}\right| \leqslant s<\frac{k m}{n}=\frac{k}{n}|A| .$$