8.5. Vasya wrote a set of different natural numbers on the board, each of which does not exceed 2023. It turned out that for any two written numbers $a$ and $b$, the number $a+b$ is not divisible by $a-b$. What is the maximum number of numbers Vasya could have written?
Solution. We will prove that the answer is $\left\lceil\frac{2023}{3}\right\rceil=675$.

Estimate. We will prove that Vasya could not write more than 675 numbers. For this, consider any three consecutive numbers $a, b=a+1$, and $c=a+2$. Note that adjacent numbers cannot be written (otherwise their difference would be one, and any number is divisible by 1). Therefore, at most only $a$ and $c$ can be written. But if they are written, their sum is $a+(a+2)=2(a+1)$, which is divisible by the difference $(a+2)-a=2$. In conclusion, out of any three consecutive numbers, at most one can be written. Thus, out of the numbers 1, 2, and 3, at most one can be written; out of 4, 5, and 6, at most one; and so on; out of 2020, 2021, and 2022, at most one. That is, out of the first 2022 numbers, at most 674 can be written, plus the number 2023, totaling 675.

Example. We will provide an example of a set of 675 numbers. Suppose Vasya wrote all numbers of the form $3n+1$ for $n$ from 0 to 674, that is, $1, 4, 7, \ldots, 2020, 2023$. Note that the sum of two such numbers has the form $(3n+1)+(3m+1)=3(n+m)+2$, which is not divisible by 3, while the difference is $(3n+1)-(3m+1)=3(n-m)$, which is divisible by 3. But a number that is not divisible by 3 cannot be divisible by a number that is divisible by 3. Therefore, the conditions of the problem are satisfied, and this set of 675 numbers is suitable.

Criteria. Only the estimate - 3 points.

Only the example with a proof that it fits - 3 points.