If they are all rational numbers, it is easy to convert the solution into a solution to the integer problem. (Just multiply all weights by the lcm of the denominators.)

But if they can be real numbers the situation is not quite so easy.

I am pretty sure that the answer in that case is still that all the weights have to be the same. Can you prove it?

]]>1.) We can redistribute the weights in a way such that the smallest amount is equal to zero (just subtract the smallest weight everywhere). If with the old weights holds that for every choice for a referee there exists a splitting of the remaining people into two teams with equal total weight, then this exists also with the new weights. And vice versa.

2.) There cannot be odd weights and even weights at the same time. Assume there are. Then if the total weight is even (odd) we can choose a person with an odd (even) weight as referee which will left the remaining people with a total odd weight (not divisible by 2). This can never be distributed into two teams with equal weight.

3.) Combining 1.) and 2.) we can assume that we have a minimal weight distribution with only even weights containing zero and fullfilling the condition. Then we can just halve every weight and get there a new distribution which fullfill also the condition. This is w.r.t. the minimality condition only possible if the initial weight was equal to zero.

Thus, all weights have to be the same, under assumption 1.) and 2.) they have to be zero.

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