$2 \cdot 26$ Try to divide the set $\{1,2, \cdots, 1989\}$ into 117 mutually disjoint subsets $A_{i}$ $(i=1,2, \cdots, 117\}$, such that
(i) each $A_{i}$ contains 17 elements;
(ii) the sum of the elements in each $A_{i}$ is the same.
[Solution] Since $1989=117 \times 17$, the set $\{1,2, \cdots, 1989\}$ can be sequentially divided into 17 segments, each containing 117 numbers. Now we appropriately place each segment of 117 numbers into $A_{1}$, $A_{2}, \cdots, A_{117}$ to satisfy condition (ii).

Starting from the fourth segment, place the numbers in even segments in ascending order into $A_{1}, A_{2}, \cdots, A_{117}$, and place the numbers in odd segments in descending order into these 117 sets. It is easy to see that the sum of 14 numbers in each set is the same. Thus, the problem reduces to how to place the first three segments of numbers $\{1,2, \cdots, 351\}$ into each set with three numbers and make the sum of the three numbers the same.

Extract the multiples of 3 from these numbers and arrange them in descending order: $\{351,348,345, \cdots, 3\}$, a total of 117 numbers. The remaining 234 numbers are arranged in ascending order and divided into two segments, each containing 117 numbers, i.e., $\{1,2,4,5,7, \cdots, 175\}$ and $\{176,178,179, \cdots, 350\}$. Place these three groups of numbers sequentially into $A_{1}, A_{2}, \cdots, A_{117}$. This satisfies the requirement. In fact, if the numbers in the second group and the third group with the same order are added, their sum is $\{177,180,183,186, \cdots, 525\}$. It can be seen that the sum of the three numbers placed in $A_{i}(i=1,2, \cdots, 117)$ is 528.