8.6. All natural numbers from 1 to 20 were divided into pairs, and the numbers in each pair were added. What is the maximum number of the ten resulting sums that can be divisible by 11? Justify your answer.
Solution: The number 11 is the only number in the set that is divisible by 11, so adding it to any other number will disrupt divisibility by 11. Therefore, all 10 sums cannot be divisible by 11. One example where nine sums are divisible by 11 is as follows:

$(1,10),(2,20),(3,19),(4,18),(5,17),(6,16),(7,15),(8,14),(9,13),(11,12)$.

Answer: 9 sums.

Recommendations for checking:

| present in the work | points |
| :--- | :--- |
| Presence of a correct example and proof of its optimality | 7 points |
| There is a proof that 10 sums cannot be obtained (in the absence of an example with 9 sums) | 3 points |
| An example is provided showing that 9 sums can be obtained (in the absence of a proof of its optimality) | 1 point |
| Correct answer without justification (with incorrect justification) | 0 points |
| Examples of a smaller number of sums (in any quantity) | not evaluated |