Can a $\Bbb Z$-module be extended to a $\Bbb Q$-module? 
Can a $\Bbb Z$-module $M$ be extended to a $\Bbb Q$-module?

It is clear that $M$ is now an abelian group.
Here $(\frac 1q+ \cdots + \frac 1q)($$q$ times)$m$ = $m$...
now from here can I arrive at some contradiction? (I guess that I can't)
 A: No, for two reasons. First, $\frac{1}{q} m$ must be a $q^{th}$ root of $m$, but such a $q^{th}$ root need not exist: for example take $M = \mathbb{Z}$. Second, if $m = qn$ then $\frac{1}{q} m = \frac{1}{q} qn = n$, so $q^{th}$ roots also need to be unique, and even when they exist in some abelian group they need not be unique: for example take $M = \mathbb{Q}/\mathbb{Z}$.
In fact an abelian group has a $\mathbb{Q}$-module structure iff each element of it is uniquely divisible, meaning all roots both exist and are unique. Since $\mathbb{Q}$-modules are $\mathbb{Q}$-vector spaces, at least assuming the axiom of choice all such abelian groups are just direct sums of copies of $\mathbb{Q}$, and most abelian groups are more interesting than this. 
On the other hand, every abelian group $M$ has a universal map into a $\mathbb{Q}$-module, namely the extension of scalars $M \otimes \mathbb{Q}$. The natural map $M \to M \otimes \mathbb{Q}$ has kernel the torsion subgroup of $M$. But even when this map is injective it is usually not surjective. 
For the experts: Tensoring the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$ with $M$ produces a Tor exact sequence which ends
$$0 \to \text{Tor}_1(M, \mathbb{Q}/\mathbb{Z}) \to M \to M \otimes \mathbb{Q} \to M \otimes \mathbb{Q}/\mathbb{Z} \to 0.$$
$\text{Tor}_1(M, \mathbb{Q}/\mathbb{Z})$ is the torsion subgroup of $M$ (this can be proven using the observation that $\text{Tor}_1(M, \frac{1}{n}\mathbb{Z}/\mathbb{Z})$ is the $n$-torsion subgroup and that $\text{Tor}$ commutes with filtered colimits), so the natural map $M \to M \otimes \mathbb{Q}$ is injective iff $M$ is torsion-free (or equivalently, flat). $M \otimes \mathbb{Q}/\mathbb{Z}$ vanishes, so the natural map $M \to M \otimes \mathbb{Q}$ is surjective, iff $M$ is divisible (or equivalently, injective). 
