Let $\mathbb F_q$ be the finite field with $q$ elements and $n>0$ be an integer.
Let's say $M \subseteq \mathbb F_q^n$ partitions into $q^{n-1}-1$ affine lines. Hence $N := \mathbb F_q^n \setminus M$ has cardinality $q$. Does this imply that $N$ is an affine line as well?
I came up with this question as I realized that this is the case for $q=3$. That is because three elements of $\mathbb F_3^n$ form an affine iff they sum up to $0$. So we have $$ \sum_{x \in N} x = \sum_{x \in F_3^n} x - \sum_{x \in M} x= 0-0=0. $$
Unfortunately this simple and beautiful idea doesn't generalize to other $q$. But from my geometric imagination I still would expect the statement to be true for arbitrary $q$.
By the way: For $q=2$ the statement is trivial since all two-element subsets form an affine line.