Answer: 400.

Solution. Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then $c-b=6$, which is impossible. Therefore, in each set of ten, no more than four numbers are selected, and in the first thousand numbers, there are no more than 400, since there are a hundred sets of ten in a thousand.

If we take all the numbers ending in $1, 2, 3$ or 4, then there will be exactly 400, but no difference will be equal to 4, 5 or 6.
2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6.