9.7 iF integers 1 to 100 are arranged in a circle in such a way that each number is either greater than or equal to its two adjacent neighbors, or less than or equal to its two adjacent neighbors. Adjacent pairs that satisfy this condition are called "good." If removing these pairs still maintains the property, how many "good" pairs can there be at a minimum?
Solve: The minimum possible number of "good" neighboring pairs is 51.

First, place the numbers 1 to 100 in a clockwise direction around a circle. Then, swap the positions of 2 and 3, 1 and 5, ..., 98 and 99. In the resulting arrangement: 1, 3, 2, 5, 4, ..., 99, 98, 100, there are exactly 51 pairs of "good" neighbors: (1, 3), (3, 2), (3, 1), (7, 6), ..., (97, 96), (99, 98), (98, 100).

Next, we prove that the number of "good" neighboring pairs is at least 51. To do this, we first prove that among any two intersecting "pairs," at least one pair is good. In fact, assume there are 5 consecutive numbers \(a_1, a_2, a_3, a_4, a_5\) such that \((a_2, a_3)\) and \((a_3, a_4)\) are not good. Therefore, \(a_1 a_2 > a_i > a_4 > a_1\). This is impossible. Thus, there are at least 50 good pairs. Furthermore, if there are exactly 50 good pairs, then "good" and "bad" neighbors must alternate. However, if we examine the numbers on either side of 100: \(a_i, a_{i+1}, 100, a_{i+2}, a_{i+3}\), it is easy to see that the pair \((a_i, a_{i+1})\) must be good because \(100 > a_i > a_{i+2} < a_{i+3}\). Similarly, the pair \((a_{i+2}, a_{i+1})\) must also be good. This indicates that "good" and "bad" neighboring pairs cannot alternate.