8,9

What is the maximum number of numbers that can be chosen from the set $1,2, \ldots, 1963$, so that the sum of no two numbers is divisible by their difference?
Let's first show that it is possible to select 655 numbers from the given set in the required manner. Indeed, let's take all numbers of the form $3k+1; k=0,1, \ldots, 654$. The difference between any two such numbers is obviously divisible by 3, while the sum of any two of them gives a remainder of 2 when divided by 3. Therefore, the sum of any two of the selected numbers does not divide their difference, i.e., the selection is made correctly. Now let's show that it is impossible to select a larger number of numbers that satisfy the conditions of the problem. Indeed, if we selected more than 655 numbers, there would be two numbers differing from each other by less than 3 (otherwise, the difference between the largest and smallest numbers in the given set would be greater than $3 \times 654=1962$, which is impossible). Therefore, there must be two numbers differing from each other by 1 or 2. In the first case, the difference between them is 1, and 1 divides any number, including their sum. In the second case, the difference between the taken numbers is 2, and therefore both these numbers have the same parity, which means their sum is an even number. In this case, the sum of the taken numbers is divisible by their difference. We have shown that it is impossible to select more than 655 numbers from the given set so that the conditions of the problem are met, and we have constructed an example of how to select 655 numbers. Therefore, the maximum number of numbers that can be selected from the given set in accordance with the conditions of the problem is 655.