It is more or less clear that if we consider a space $\mathbb{F}_2^k$ and pick $n$ uniformly random vectors then with a constant probability any ${2 \over 3} k$-element subset of this random set is linearly independent.
Although it seems like there is an example such that any $k$-element subset is linearly independent. Does anyone aware of such a collection of vectors? (not necessarily for $\mathbb{F}_2$, but it's preferable).
In a finite field like $\Bbb{F}_2$, it will depend on $n$. Let's not forget that $\Bbb{F}_2^k$ is a finite vector space, containing only $2^k$ vectors. Obviously, if we had a set of $n = 2^k$ vectors, then it would be all of $\Bbb{F}_2^k$, which contains plenty of linearly dependent subsets of $k$ vectors.
On the other hand, if we have an infinite field $F$, then we actually can find such a set for any $n$. All we do is let $r_1, \ldots, r_n$ be distinct elements of $F$. Then the vectors $$v_i = (1, r_i, r_i^2, r_i^3, \ldots, r_i^{k-1}),$$ for $i = 1, \ldots, n$ form such a set. If we take any $k$ of them and put them in rows of a matrix, then the result is a Vandermonde matrix, which are invertible if and only if the ratios for each row are distinct. Thus, any selection of $k$ vectors from the set will be linearly independent.