Given a natural number $k\geq 1$. Is there always a odd natural number $n>k$ , so that for any k pairwise different boolean vectors $v_1 , v_2 ,\ldots, v_k\in \mathbb{Z}_2^n$ with Hamming distance $d(v_i,v_j)< n$ for all $i \neq j$ there is a vector $w\in\mathbb{Z}_2^n$ with $0<d(w_,v_i)\leq \frac{n-1}{2}$ for all $i$ ?
I have verified this for $k\leq 4$, but can't generalize the proof. Any tips welcome.
Found a counterexample for $k=4$ myself.
$v_1=(1,0,1,1,1,\ldots,1)$
$v_2=(1,1,0,0,1,\ldots,1)$
$v_3=(1,0,1,1,0,\ldots,0)$
$v_4=(1,1,0,0,0,\ldots,0)$
Then there is no $w\in\mathbb{Z}_2^n$ with $d(w,v_i)\leq \frac{n-1}{2}$ for all $i$.
We can write those as bitstrings:
$ v_1=1abc1^{n-4} $
$ v_2=1\overline{abc}1^{n-4} $
$ v_3=1abc0^{n-4} $
$ v_4=1\overline{abc}0^{n-4} $
where $abc=011$ and $\overline{abc}=100$ is the complement.
Now any $w$ must have either $\geq \frac{n-3}{2}$ one's or $\geq \frac{n-3}{2}$ zero's in the last $n-4$ positions.
Assume it has $\geq \frac{n-3}{2}$ one's in the last $n-4$ positions. Then it differs from $v_3$ and $v_4$ at $\frac{n-3}{2}$ positions in the $0^{n-4}$ suffix.
But it must also differ either from $abc$ or $\overline{abc}$ in at least 2 positions. So $d(w,v_3)\geq \frac{n-3}{2}+2=\frac{n+1}{2}>\frac{n-1}{2}$ or $d(w,v_4)>\frac{n-1}{2}$.
Analogously for the other case.