(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,
if n is lucky, so is 2n+2.You can double each of the numbers in n's lucky list, so they sum to 2n and the recipricals sum to 1/2, then add '2' so they sum to 2n+2 and the recipricals sum to 1.
To address the odd numbers, use this trick:
if n is lucky, and its “lucky list” contains a '2', then n+7 is lucky.You 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 number n is lucky by using the first trick on n-7, then using the second trick.
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).