8. (10 points) From 1 to 1000, the maximum number of numbers that can be selected such that the difference between any two of these numbers does not divide their sum is $\qquad$.
【Analysis】To get the maximum number, the divisor must be as small as possible. Since natural numbers can be divided into 2 categories based on the remainder when divided by 2, i.e., remainders are: 0, 1, the difference between any two numbers in these categories can divide their sum, which does not meet the requirement; then consider 3, natural numbers can be divided into 3 categories based on the remainder when divided by 3, i.e., remainders are: $0, 1, 2$, then further divide the remainders into two categories, 1 and 0 and 2, to derive the answer.

【Solution】Clearly, natural numbers can be divided into 3 categories based on the remainder when divided by 3, i.e., remainders are: $0, 1, 2$. All numbers that leave a remainder of 1 when divided by 3, the sum of any two such numbers leaves a remainder of 2 when divided by 3, and their difference can be divided by 3, which meets the requirement. The same applies to all numbers that leave a remainder of 2 when divided by 3, i.e., $2+2=4, 4 \div 3$ still leaves a remainder of 1. In the range from 1 to 1000, there are 334 numbers that leave a remainder of 1 when divided by 3, and 333 numbers that leave a remainder of $0, 2$. Therefore, taking the 334 numbers that leave a remainder of 1 when divided by 3, these numbers meet the requirement; hence the answer is: 334.