There must be some extra condition on the size of the integers in the set. Otherwise, you could have the numbers . In any set of four of those numbers, the largest one would swamp the sum or difference of the other three. So a–b+c–d would be much larger than 2009, and could not be a factor of it.

If you want to avoid pairs that differ by 3, then in the set 1,4,7,10,13,16,19,22,25,28 you could not pick any two consecutive elements. So the maximum number chosen from this set would be 5. Similarly for the sets 2,5,8,...,29 and 3,6,9,...,30, giving a total of at most 15 numbers altogether.