If n is any non-zero integer and G is a topological abelian group, under what conditions is the continuous homomorphism $g \mapsto ng$ open or weakly open?
If this map happens to be open for all n and at the same time, G is divisible and torsion-free, therefore admitting a vector space structure over $\mathbb{Q}$, can we affirm that G becomes a topological vector space over the rationals?
This is not true even for torsion-free divisible groups. Consider $\mathbb{Q}$ with the topology where $\mathbb{Z}$ is an open subgroup and $\mathbb{Z}$ has the $2$-adic topology. Then the sequence $(2^k)$ converges to $0$ but $(\frac{1}{n}2^k)$ does not converge to $0$ unless $n$ is a power of $2$, so multiplication by $n$ is not open unless $n$ is a power of $2$.
Moreover, even if $G$ is torsion-free and divisible and multiplication by $n$ is open for all $n$, then $G$ does not have to be a topological vector space over $\mathbb{Q}$ (with its usual topology). For instance, $G$ could be $\mathbb{Q}$ with the discrete topology, and then scalar multiplication is not continuous. (Of course, scalar multiplication will always be continuous if you give the scalar field the discrete topology, but I assume you wanted the usual topology on $\mathbb{Q}$.)