Clearly it is possible for such sets to exist even if no combination of the weights themselves can be balanced against any combination of the remaining weights - the set {1,2,4,8..} is just one such example.

On the other hand, the set {1,1,2,4} is also 'self-measuring', because, assuming that one of the weights were unmarked, a stranger could neverthless establish its value by weighing it in the scales with the others.

Question: You are allowed 5 weights. They must form a set which is complete, and also self-measuring.

What is the largest possible value of W, given these constraints?

Bonus: What is the largest possible value of W, if 8 weights are allowed?

I kept thinking about that self-measuring part, kept thinking about it, kept thinking about it, completely forgot to account for it in my solution. So, unless your Mathematica code can find better, here are my new sequences, based on my already erased previous message:

As {1, 3, 9, 27, 81} is optimized for the sums and differences only, and NOT self-measuring, let us remove the last term. This leaves {1, 3, 9, 27}, which is the optimal set of size 4, for forming sums and differences. Now, the fifth number to be added will be the sum of those numbers (1 + 3 + 9 + 27 = 40). That way, we can use the first 4 numbers to represent anything from 1 to 40, and add 40 to those values for anything from 41 to 80. In fact, any set {x, 3, 9, 27, x+39} with 5 distinct positive integers, and x not being a multiple of 3 seems to result in W=80.