11.4. One hundred and one non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of negative numbers that can be among these 101 written numbers?
Answer: 67.

Solution: Consider any 3 consecutive numbers. All of them cannot be negative. Among them, there is a positive one. Fix this positive number and its neighbor (this number can be negative), and divide the remaining 99 numbers into 33 groups of 3 consecutive numbers. In each such group, there will be no more than two negative numbers. Thus, the total number of negative numbers is no more than $1 + 2 \times 33 = 67$. Such a situation is possible. Number the numbers in a circle. Let the positive numbers be those with indices $1, 4, 7, 10$, ..., 100, and the rest be negative. It is easy to verify that if all positive numbers are equal to 5, and all negative numbers are equal to -2, then each number is greater than the product of the two numbers following it in the clockwise direction.

Comment: Correct answer without justification - 0 points.

Proved that the number of negative numbers is no more than 67 - 4 points.

Provided an example with 67 negative numbers - 3 points.