7. Variant 1.

103 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there are at least two even numbers. What is the minimum number of even numbers that can be in the entire circle?
Answer: 42.

Solution.

We will show that there will be 3 consecutive numbers, among which there are at least 2 even numbers. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them there are at least 6 even numbers. But these 15 numbers can also be divided into 5 sets of 3 consecutive numbers. Therefore, by the Pigeonhole Principle, in one of these triplets there are at least 2 even numbers.

Let's fix these 3 numbers. Among them, there are at least 2 even numbers. The remaining 100 numbers can be divided into 20 sets of 5 consecutive numbers. In each such set, there will be at least 2 even numbers. Thus, the total number of even numbers is at least $2 + 2 \cdot 20 = 42$. Such a situation is possible. We can number the numbers in a circle. The even numbers can be those with numbers $1, 2, 6, 7, \ldots, 96, 97, 101, 102$. (Other examples are also possible.)