11.3 A weirdo chose 677 different natural numbers from the list $1,2,3, \ldots, 2022$. He claims that the sum of no two of the chosen numbers is divisible by 6. Did he go too far with his claim?
Solution: Suppose the eccentric is right. From the list specified in the condition, there are exactly 377 remainders of division by 6 of each type from 0 to 5. Numbers with remainders 0 and 3 can be taken no more than one each. It is impossible to take numbers with remainders 1 and 5 simultaneously, so in the eccentric's set, there are either those with a remainder of 1 or 5, and therefore no more than 337 numbers with a remainder of 1 or 5. Similarly, with a remainder of 2 or 4, there are no more than 337 numbers in total. In total, to ensure that the sum of no two numbers is divisible by 6, the eccentric could choose no more than $1+1+337+337=676$ numbers. Contradiction.

## Criteria:

- It is noted that no more than one number divisible by 3 and no more than one number with a remainder of 3 can be taken - 1 point.