*6. From $1,2, \cdots, 1996$, select $k$ numbers such that the sum of any two numbers cannot be divisible by their difference. Then the maximum possible value of $k$ is $\qquad$ .
6. 666 .
Classify $1,2, \cdots, 1996$ by their remainder when divided by 3:
$$
\begin{array}{l}
3 k+1: 1,4,7, \cdots, 1996, \text { a total of 666; } \\
3 k+2: 2,5,8, \cdots, 1994, \text { a total of 665; } \\
3 k: 3,6,9, \cdots, 1995, \text { a total of 665. }
\end{array}
$$

Select all $3 k+1$ numbers, a total of 666, where the sum of any two is $3 k+2$, which cannot be divided by the difference $3 k$, meeting the problem's requirements.

If 667 numbers can be selected, consider $(1,2,3),(4,5,6), \cdots$, $(1993,1994,1995),(1996)$ as 666 drawers. There must be two numbers in the same drawer. Their difference $x-b$ is less than 3.
If $a-b=1$, then $a-b \mid a+b$;
If $a-b=2$, then $a$ and $b$ have the same parity.
Thus, $a-b \mid a+b$, so the maximum number that can be selected is 666.