Let $v_1, \dotsc, v_N$ be a set of vectors of $\mathbb{F}_2^n$ which has the following property: for any choice of $1 \leq j_1 < \dotsb < j_n \leq N$, the vectors $v_{j_1}, \dotsc, v_{j_n}$ are linearly independent.
What is the maximum possible $N$? Is there an explicit choice of $v_1, \dotsc, v_N$ in that case?
Thanks!
The result is the somewhat uninteresting upper bound $N\le n+1$. This is relatively well-known in a coding theoretical setting. Let me give a simple argument.
The vectors $v_1,v_2,\ldots,v_n$ obviously must form a basis of $\Bbb{F}_2^n$. Thus we can start using them as a basis for the coordinate system, and w.l.o.g. assume that $v_1=(1,0,\ldots,0)$, $v_2=(0,1,0,\ldots,0)$, $\ldots$, $v_n=(0,\ldots,0,1)$. Let $k>n$ and let $v_k=(b_1,b_2,\ldots,b_n)$. If some $b_i=0$, then the set $v_1,v_2,\ldots,v_{i-1},v_{i+1},v_{i+2},\ldots,v_n,v_k$ is linearly dependent. Therefore we can conclude that $v_k=(1,1,\ldots,1)$ for all $k, n<k$. There is only one such vector, so we must have $k\le n+1$ and therefore also $N\le n+1$.
It is easy to see that the set $v_1,v_2,\ldots,v_n,v_{n+1}=\sum_{i=1}^nv_i$ has the prescribed property. Therefore $N=n+1$ is achieved.