12 From the set $M=\{1,2, \cdots, 2008\}$ of the first 2008 positive integers, select a $k$-element subset $A$, such that the sum of any two numbers in $A$ cannot be divisible by their difference. Then the maximum value of $k$ is $\qquad$.
(12) First, we can take the set $A=\{1,4,7, \cdots, 2008\}$, where the sum of any two numbers in $A$ cannot be divisible by 3, while their difference is a multiple of 3; second, we divide the numbers in $M$ from smallest to largest into 670 segments of three numbers each:
$$
1,2,3,|4,5,6,| 7,8,9,|\cdots,| 2005,2006,2007, \mid 2008 .
$$

If we select 671 numbers from $A$, there must be two numbers $x$ and $y$ from the same segment, then $|x-y|=1$ or 2. Note that $x-y$ and $x+y$ have the same parity, so $(x-y) \mid (x+y)$. Therefore, the maximum value of $k$ is 670.