Let $V$ be a vector space over $\mathbb{Z}/2\mathbb{Z}$. Can there be a set $S$ of $2 n$ vectors in $V$ such that any $n$ vectors in $S$ span a space of dimension exactly $n-1$, but no $n$ vectors $v\_1,\dotsc ,v\_n\in S$ satisfy $v\_n = v\_1 + v\_2 + \dotsc + v\_{n-1}$?



(Barely) linearly independent vectors over $\mathbb{Z}/2\mathbb{Z}$
Since every $n$ vectors span a space of dimension $n-1$, the whole set $S$ spans a space of dimension $n-1$, and we can assume that $V$ has dimension $n-1$. 
Choose $n-1$ linearly independent vectors $T=\{t\_1,t\_2,\dots,t\_{n-1}\}\subseteq S$ and assign the standard basis to them, i.e. $t\_k=(0,0,\dots,0,1,0,\dots,0)$ where the $1$ is at index $k$. 


Let $x,y\in S\setminus T$ be distinct. For each $k\in\{1,\dots,n-1\}$, $x\_k\neq 0$ or $y\_k\neq 0$, otherwise there are $n$ vectors in $S$ spanning a $n-2$ dimensional space: $x$, $y$, and all the vectors in $T$ excluding $t\_k$.


By the pigeonhole principle, there are at most $n-1$ zero indices in $S\setminus T$. So there are at most $n$ vectors in $S\setminus T$ (otherwise there would be two all-$1$ vectors). 


So the size of $S$ is at most $2n-1$. As a result, there is no set $S$ of size $|S|=2n$ such that any $n$ vectors from $S$ span an $(n-1)$-dimensional subspace.


