*(We'll assume the puzzle was asking about *positive* integers. If negative integers
are allowed, I believe all numbers are lucky.)*

There are **thirteen** unlucky numbers: 2, 3, 5, 6, 7, 8, 12, 13, 14, 15, 19, 21, 23.

After that, there are increasingly many combinations of integers, and all numbers are lucky. To prove it, use two simple tricks. First,

ifYou can double each of the numbers innis lucky, so is2n+2.

To address the odd numbers, use this trick:

ifYou can trade a 2 for a 3 and a 6. 1/2 is the same as 1/3+1/6, and replacing 2 with 3+6 increases the sum by 7. This works together with the previous trick. You can show that an odd numbernis lucky, and its “lucky list” contains a '2', thenn+7is lucky.

So, finally, if we can
establish a lucky streak
from `n` to `2n+2+7`, then we know we can apply these
two tricks *ad infinitum*, thus showing that all integers with greater value are also lucky.
The first such streak, which can be shown with a comination of the tricks above and brute force, is:

24 (4, 6, 12, 2),

25 (5, 5, 5, 5, 5),

26 (6, 6, 6, 4, 4),

27 (6, 6, 6, 3, 6),

28 (10, 5, 5, 4, 4),

29 (8, 8, 4, 3, 6),

30 (15, 10, 3, 2),

31 (4, 6, 12, 3, 6),

32 (18, 9, 3, 2),

33 (15, 5, 5, 5, 3),

34 (8, 8, 8, 8, 2),

35 (12, 12, 4, 4, 3),

36 (8, 8, 6, 12, 2),

37 (15, 10, 3, 3, 6),

38 (12, 12, 6, 6, 2),

39 (18, 9, 3, 3, 6),

40 (12, 6, 6, 6, 6, 4),

41 (8, 8, 8, 8, 3, 6),

42 (12, 12, 12, 4, 2),

43 (8, 8, 6, 12, 3, 6),

44 (24, 8, 6, 3, 3),

45 (12, 12, 6, 6, 3, 6),

46 (16, 16, 8, 4, 2),

47 (20, 6, 6, 6, 5, 4),

48 (18, 9, 6, 6, 6, 3),

49 (12, 12, 12, 4, 3, 6),

50 (8, 12, 24, 4, 2),

51 (30, 10, 5, 3, 3),

52 (10, 10, 10, 10, 10, 2),

53 (16, 16, 8, 4, 3, 6),

54 (12, 12, 12, 8, 8, 2),

55 (28, 14, 7, 4, 2),

56 (12, 12, 12, 6, 12, 2),

57 (8, 12, 24, 4, 3, 6).