$2 \cdot 12$ Let $S=\{1,2, \cdots, 1963\}$, how many numbers can be selected from $S$ at most, so that the sum of any two numbers cannot be divisible by their difference?
［Solution］Let
$$
T=\{3 k+1 \mid k=0,1, \cdots, 654\},
$$

then $T \subset S$ and $|T|=655$. Any two numbers in $T$ have a difference that is a multiple of 3, but their sum is not a multiple of 3. Therefore, the sum of any two numbers in $T$ cannot be divisible by their difference. Thus, the maximum value we seek is at least 655.

For any 656-element subset $M$ of $S$, there must be two numbers whose difference is no more than 2. Let these two numbers be $a$ and $b$ with $a > b$. If $a-b=1$, then it is clear that $(a-b) \mid (a+b)$; if $a-b=2$, then $a+b$ is also even, and thus $(a-b) \mid (a+b)$. This shows that any subset with more than 655 elements does not satisfy the condition given in the problem.

In conclusion, the maximum number of numbers that can be selected such that the sum of any two numbers is not divisible by their difference is 655.