This is more or less a by-product in my search of non-measurable sets, aside from the well-known $\mathbb{R}/\mathbb{Q}$ example. Maybe it's interesting, maybe not; who knows. Anyway, here is the precise description. We assume the Axiom of Choice.
Choose $B$ to be a Hamel basis of the $\mathbb{Q}$-vector space $\mathbb{R}$. It is known that $B$ is either non-measurable or of measure zero. Now, to muddy the waters, we investigate the linear combinations. Define $A_n$ as the set of all $n$-combinations of $B$.
Equivalently, $A_n = \{ x\in \mathbb{R} \mid x = \lambda_1 b_1 + \cdots + \lambda_n b_n, \lambda_i \in \mathbb{Q}, b_i\in B\}$, where $n$ is any positive integer. (You may stipulate $A_0 = \{0\}$ if you like.)
With Steinhaus Theorem we can easily see $A_n$ is either measure-zero or non-measurable. Moreover, they form an ascending sequence of sets. Also, we observe that it can't be the case if every $A_n$ has zero measure, because $\mathbb{R} = \bigcup_n A_n$. In other words, each real number can be (uniquely) written as a rational combination of finite elements in $B$. So there must be a minimal integer $N(B)$, such that all $A_n$'s are non-measurable once $n$ goes beyond $N(B)$.
I don't know if $N(B)$ is worth studying; I'm even not sure what are right questions to bring up. Nonetheless, at least it gives some information on the complexity of $B$ in the context of Lebesgue measure theory. In a related question, we see both non-measurable and zero-measure instances of $B$ can be constructed. If $B$ is itself non-measurable, then naturally $N(B)$ is $1$. In this case $B$ is very nasty. On the other hand, a zero-measure $B$ doesn't reveal all its complexity: $A_1$ is indeed zero-measure, but higher $A_n$ can be a total mess.
One question of mine is, could every possible $N(B)$ be reached? I shall leave other interesting questions to experts to fill in.
