= Solution
Let $T$ consist of all <torsion elements of a module> in $M$. If $rm=0$ and $sn=0$ with $r,s\ne0$, then $rs(m+n)=0$. Since $R$ is an <integral domain>, $rs\ne0$. 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 $r\ne0$, then $rm\in T$, so some $s\ne0$ satisfies $srm=0$. The domain property again gives $sr\ne0$, 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.
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