8.5. One hundred and one numbers are written in a circle. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 101 written numbers?
Answer: 41.

Solution. Consider any 5 consecutive numbers. Among them, there is a positive one. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there will be at least two positive numbers. Thus, the total number of positive numbers is at least $1+2 \cdot 20=41$. Such a situation is possible. Number the numbers in a circle. The positive ones can be those with numbers $1,3,6,8,11, \ldots, 98,101$.

Comment. Correct answer without justification - 0 points.

Proved that the number of positive numbers is not less than 41 - 5 points. Provided an example with 41 positive numbers - 2 points.