7. (50 points) From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property?
Answer: 40.

Solution. Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this group. Suppose there is another number $C$, and $j=C \bmod 5$. Then the numbers $i+j$ and $5-i+j$ are divisible by 5, as well as their difference, which is $2 i-5$, which is impossible for $i \neq 0$.

Now consider the alternative case, where all numbers in the group are divisible by 5. In the given range, there are exactly 40 numbers that are multiples of 5, and all these numbers can be included in the group simultaneously.