Lemma: If V is maximal subset then V consists of all vectors in some half-plane, and doesn't contain any other vector. Proof: Let $\vec{s}$ be vector that is sum of all vectors in V. Consider half-plane which is determined by line perpendicular to $\vec{s}$ at the origin and which contains $\vec{s}$. First, notice that each vector from the half-plane is contained in V. If some vector $\vec{v}$ isn't, then we can see that $ |\vec{s} + \vec{v}| > |\vec{s}| $ - just consider triangle which consists of $ \vec{s}, \vec{v}, \vec{s} + \vec{v} $ and observe obtuse angle across the side $ \vec{s} + \vec{v} $. Same way, one can prove there is no vector which is element of V and that is not in the half-plane. So lemma is proven. Now, all "interesting" half-planes are determined with one vector and direction (clockwise or counter-clockwise), so there are at most $ 2n $ maximal subsets.