11. (40 points) The numbers from 1 to 600 are divided into several groups. It is known that if a group contains more than one number, then the sum of any two numbers in this group is divisible by 6. What is the minimum number of groups that can be formed?
Answer: 202.

Solution: For $k=0,1, \ldots, 5$, let $G_{k}$ be the set of numbers from 1 to 600 that give a remainder of $k$ when divided by 6. Suppose a number $a$ from $G_{k}$ is included in some group. If another number $b$ is also in this group, then it belongs to $G_{6-k}$, otherwise $a+b$ is not divisible by 6. Suppose there is a third number $c$ in this group. Then it lies in both $G_{k}$ and $G_{6-k}$. This is possible only for $k=0$ and $k=3$. Numbers not in $G_{0}$ and $G_{3}$ can only be included in two-element groups, and there are 400 such numbers. Therefore, they form at least 200 groups. The sets $G_{0}$ and $G_{3}$ can form groups, but they cannot be combined into one. Thus, at least 202 groups are required in total.

We will provide an implementation of the partition into 202 groups. One of them will be $G_{0}$, another will be $G_{3}$, and we will also choose groups of the form $\{6 n+k, 6 n+6-k\}$, where $n=0,1, \ldots, 99$ and $k=1,2$.