Example 5 (International Mathematical Olympiad) Try to divide the set $\{1,2, \cdots, 1989\}$ into 117 mutually disjoint subsets $A_{i}$ $(i=1,2, \cdots, 117)$, such that:
(1) Each $A_{i}$ contains 17 elements;
(2) The sum of the elements in each $A_{i}$ is the same.
Analysis Since $1989=117 \times 17$, the set $\{1,2, \cdots, 1989\}$ can be sequentially divided into 17 segments, each containing 117 numbers. Clearly, if the 117 numbers in each segment are appropriately placed into $A_{1}, A_{2}, \cdots, A_{11}$ to satisfy condition (2), the problem is solved.
Solution The numbers in the set $\{1,2, \cdots, 1989\}$ are sequentially divided into 17 segments, each containing 117 numbers.
Starting from the 4th segment, the numbers in the even-numbered segments are placed in $A_{1}, A_{2}, \cdots, A_{117}$ in ascending order, and the numbers in the odd-numbered segments are placed in these 117 subsets in ascending order. It is easy to see that the sum of 14 numbers in each set is equal. Thus, the problem reduces to how to place the first three segments of numbers $\{1,2, \cdots, 351\}$ into each subset in groups of 3, such that the sum of the 3 numbers in each group is equal.

Extract the multiples of 3 from these numbers and arrange them in ascending order: $\{351,348,345, \cdots, 6,3\}$, a total of 117 numbers, and place them sequentially into $A_{1}, A_{2}, \cdots, A_{117}$. The remaining 234 numbers are arranged in ascending order and divided into two segments, each containing 117 numbers, i.e., $\{1,2,4,5,7, \cdots, 173,175\}$ and $\{176,178,179, \cdots, 349,350\}$. Place these two segments of numbers sequentially into $A_{1}$, $A_{2}, \cdots, A_{117}$ to meet the requirements. In fact, if the numbers in these two segments are added in sequence, their sums are $\{177,180, 183,186, \cdots, 522,525\}$. It can be seen that the sum of the 3 numbers placed in each $A_{i}$ is 528.