3. (20 points) The pensioners of one of the planets in the Alpha Centauri system enjoy spending their free time solving cybrow solitaire puzzles: they choose natural numbers from a certain interval $[A, B]$ in such a way that the sum of any two of the chosen numbers is not divisible by a given number $N$. Last week, the newspaper "Alphacentauri Panorama" offered its readers a solitaire puzzle with the values $A=1353, B=2134, N=11$. What is the maximum number of numbers that can be a solution to such a solitaire puzzle?
Answer: 356.

Solution: For $k=0,1, \ldots, 10$, let $I_{k}$ be the set of all numbers in $[A, B)$ that give a remainder of $k$ when divided by 11. Since $A$ and $B$ are multiples of 11, all sets $I_{k}$ contain an equal number of elements. Therefore, all numbers in $[A, B)$ that are not multiples of 11 can be paired as $(x, y)$, where $x \in I_{k}, y \in I_{11-k}$ for some $k$ from $\{1, \ldots, 5\}$. The number of such pairs is $\frac{2134-1353}{11} \cdot 5=355$. Clearly, from each pair, at most one number can be included in the final set. Additionally, this set can contain at most one number that is a multiple of 11. Thus, the solution includes no more than 356 numbers.

Now, we will show that a set of 356 numbers is possible. Include $A$ and all numbers that give an odd remainder when divided by 11 in this set. Take two distinct numbers $x$ and $y$ from this set, different from each other and from $A$. Then $(x \bmod 11) + (y \bmod 11)$ is an even number between 2 and 18. Therefore, it is not divisible by 11, and thus $x+y$ is also not divisible by 11. Clearly, $A+x$ is also not
divisible by 11. This means that such a set satisfies the problem's conditions. It remains to note that this set contains $\frac{2134-1353}{11} \cdot 5 + 1 = 356$ numbers.