Smallest size of set of real numbers such that the sum of any seven is strictly positive, and the sum of any eleven is strictly negative So I just got asked a question that riddled me. 

If you have a set of real numbers, such that the sum of any 7 numbers from this set is strictly positive and the sum of any 11 numbers from this set is strictly negative, then what is the smallest possible size of this set? 

I've managed to prove that the size can't be 11 but beyond that I'm bamboozled. (I'm not even sure if this is possible, I was asked this question in an interview)
 A: Are you sure that is the correct version of the question? The simple answer seems to be 0, as for the empty set both conditions are fullfilled (if no 7/11-element subsets exists, than making any statement about all of them will always be true).
Note that this seems to be related/derived from to a question from 1977 International Mathematical Olympiad:

In a finite sequence of real numbers the sum of any seven successive
  terms is negative, and the sum of any eleven successive terms is
  positive. Determine the maximum number of terms in the sequence.

Notice however the important differences: Sequence instead of set and the maximum number is sought instead of the minimum. The positive/negative exchange seems to be unimportant, one can just exchange the sign of any set/sequence member to get from one formulation to the other.
A: It is rather easy to prove that the conditions are impossible to meet. Imagine we have a sequence where any set of 11 numbers is negative. We could always simply remove the four highest numbers from the problem, which guarantees that the solution is still negative. If all the removed numbers are positive, then the total must have decreased, so it must still be negative. If, however, a negative number is removed, it must mean that no positive numbers are left, and therefore, the total is negative.
