$v_1,v_2,\dots,v_n\in \mathbf R^2$ are $n$ 2-dim vectors of length at least 1.
Show that we can choose at most $m={n\choose \lfloor n/2\rfloor}$ subsets of $\{v_1,v_2,\dots,v_n\}$ , say $S_1,S_2,\dots,S_m$, such that $\forall i,j\in [m], |f(S_i)-f(S_j)|<1$,
where $f(S)=\sum_{v\in S}v$ (i.e. the sum of all its vectors).
For example, when $v_1=(0,1),v_2=(0,-1),v_3=(1,0),v_4=(-1,0)$, at most 4 subsets can be chosen to meet the conditions if I'm right. They are: $\emptyset, \{v_1,v_3\},\{v_2,v_4\},\{v_1,v_2,v_3,v_4\}$. We cannot choose ${4\choose 2}=6$ subsets anyway.
I have no idea how to solve this problem as I am not sure how to say "$|f(S_i)-f(S_j)|<1$" more mathematically.
Thanks in advance.