9.5. One hundred integers are written in a circle. Each number is greater than the sum of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among the written ones

(S. Berlov)
Answer: 49.

Solution: Suppose that two non-negative numbers stand next to each other. Then the number preceding them is greater than their sum, meaning it is positive. Similarly, the number before it is also positive, and so on. In the end, we get that all numbers are non-negative; but then the smallest of them cannot be greater than the sum of the next two - a contradiction.

Thus, among any two consecutive numbers, there is at least one negative number. This means there are no more than 50 positive numbers. Suppose there are exactly 50, then they alternate with negative numbers. Now consider three consecutive numbers $-a, b, -c$ (here $a, b, c > 0$). Then $-a > b - c > -c$, meaning any negative number is strictly greater than the next negative number. Since the numbers are arranged in a circle, this is impossible. Therefore, there are no more than 49 positive numbers.

It remains to provide an example where there are exactly 49 positive numbers. For instance, the following arrangement works:

$-200, 1, -202, 1, -204, 1, -206, 1, \ldots, -296, 1, -298, -99$.