6. The natural number $k$ satisfies the following property: in $1,2, \cdots, 1988$, $k$ different numbers can be selected such that the sum of any two of these numbers is not divisible by their difference. Find the maximum value of $k$. (26th Mo

将上面的文本翻译成英文，保留了源文本的换行和格式。
6. The maximum value of $k$ is 663. First, we prove that $k \leqslant 663$. We note the following fact: when $x - y = 1$ or 2, we have $x - y \mid x + y$. This implies that in any sequence of 3 consecutive natural numbers, if we take any two numbers $x$ and $y$, then $x - y \mid x + y$. If $k > 663$, let $A_{i} = \{3i - 2, 3i - 1, 3i\} (i = 1, 2, \cdots, 662)$, and $A_{663} = \{1987, 1988\}$. When we select $k \geqslant 664$ numbers and place them into these 663 sets, at least one set will contain two numbers $x$ and $y$. In this case, it is clear that $x - y \mid x + y$, leading to a contradiction. 

Secondly, when $k = 663$, let $A = \{1, 4, 7, \cdots, 1987\}$. For any two numbers $a_{i}$ and $a_{j}$ in $A$, we have $a_{i} - a_{j} = (3i - 2) - (3j - 2) = 3(i - j)$, and $a_{i} + a_{j} = (3i - 2) + (3j - 2) = 3(i + j - 1) - 1$. Therefore, $a_{i} - a_{j}$ does not divide $a_{i} + a_{j}$.