$2 \cdot 89$ 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 all $A_{i}$ is the same.
[Solution 1] Since $1989=117 \times 17$, the set $\{1,2, \cdots, 1989\}$ can be sequentially divided into 17 segments, each containing 117 numbers. Clearly, as long as the 117 numbers in each segment are appropriately placed into $A_{1}$, $A_{2}, \cdots, A_{117}$ to satisfy condition (2), the problem is solved.

Starting from the 4th segment, place the even-numbered segments in increasing order into $A_{1}, A_{2}, \cdots, A_{117}$ and place the odd-numbered segments in decreasing order into these 117 subsets. It is easy to see that the sum of 14 numbers in all sets is equal. Thus, the problem reduces to how to place the first three segments of numbers $\{1,2, \cdots, 351\}$ into each set with 3 numbers such that the sum of the 3 numbers is equal.

Extract the multiples of 3 from these numbers and arrange them in descending order: $\{351,348,345, \cdots, 6, 3\}$, 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., $\{1,2,4,5,7, \cdots, 173,175\}$ 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 sequentially, 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.

[Solution 2] As in Solution 1, divide $\{1,2, \cdots, 1989\}$ sequentially into 17 segments, each containing 117 consecutive natural numbers. Place the first two segments of the last 4 segments in increasing order into the sets $A_{1}$, $A_{2}, \cdots, A_{117}$, and place the last two segments in decreasing order into the sets $A_{1}$, $A_{2}, \cdots, A_{117}$. Clearly, the sum of the 4 numbers placed in all sets is equal. Thus, the problem is reduced to dividing $\{1,2, \cdots, 1521\}$ into 117 mutually disjoint subsets, each containing 13 elements and the sum of the elements in all subsets being equal.

Divide the 117 sets $A_{1}, A_{2}, \cdots, A_{117}$ sequentially into 13 groups, each containing 9 sets and denote these groups as $B_{1}, B_{2}, \cdots, B_{13}$. Divide the 117 numbers in each of the 13 segments into 13 groups, each containing 9 consecutive natural numbers. For each $k, 1 \leqslant k \leqslant 13$, place the 13 groups of numbers from the $k$-th segment sequentially into the set groups $B_{k}$, $B_{k+1}, \cdots, B_{13}, B_{1}, B_{2}, \cdots, B_{k-1}$. It is easy to see that the sum of all numbers in each set group is equal. Thus, the remaining problem is how to place the 13 groups of numbers in each set group appropriately into the 9 sets in the group.

Assume the 9 sets are $A_{1}, A_{2}, \cdots, A_{9}$. Place the first two groups of the last 4 groups of 13 groups of numbers in increasing order into $A_{1}, A_{2}, \cdots, A_{9}$, and place the last two groups in decreasing order into $A_{1}, A_{2}, \cdots, A_{9}$. Thus, the sum of the 4 numbers placed in each set is equal. Finally, place the $j$-th group of the remaining 9 groups of numbers in increasing order into the sets $A_{j}, A_{j+1}, \cdots, A_{9}$, $A_{1}, \cdots, A_{j-1}$, $j=1,2, \cdots, 9$. It is easy to see that the sum of all numbers in each of the 9 sets is equal.