 Solution

Let $T$ consist of all torsion elements of a module in $M$. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$. Since $R$ is an integral domain, $rs\ne 0$. Also $r(am)=0$ for every scalar $a\in R$. Thus $T$ is the torsion submodule.

If $r(m+T)=0$ in $M/T$ with $r\ne 0$, then $rm\in T$, so some $s\ne 0$ satisfies $srm=0$. The domain property again gives $sr\ne 0$, hence $m\in T$. Therefore $M/T$ is a torsion-free module.

Finally, any submodule $N$ with the stated properties satisfies $N\subseteq T$. Conversely, if $m\in T$, its coset in $M/N$ is torsion and therefore zero, so $m\in N$. Hence $\boxed{N=T}$ is the unique choice.