Example 4: From the numbers $1, 2, \cdots, 2012$, select a set of numbers such that the sum of any two numbers cannot be divisible by their difference. How many such numbers can be selected at most?
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities)
Solve: Divide $1,2, \cdots, 2012$ into
$$
\begin{array}{l}
(1,2,3),(4,5,6), \cdots, \\
(2008,2009,2010),(2011,2012)
\end{array}
$$

these 671 groups.
If at least 672 numbers are taken, then by the pigeonhole principle, there must be two numbers in the same group, let's say $a$ and $b$ ($a>b$).
Thus, $a-b=1$ or 2.
When $a-b=1$, $(a-b) \mid (a+b)$, which is a contradiction.
When $a-b=2$, $a$ and $b$ have the same parity, thus, $a+b$ is even. Therefore, $(a-b) \mid (a+b)$, which is a contradiction.
Therefore, the maximum number of numbers that can be taken is 671.
For example, taking $1,4,6, \cdots, 2011$ these 671 numbers satisfies the condition.