= Solution
Write $\alpha=a/q+\beta$, where $|\beta|\leq q^{-2}$. Split the $m$-interval into $O(M/q+1)$ consecutive blocks of length at most $q/2$. If distinct integers $m,n$ lie in one block, then $0<|m-n|<q/2$. Since $(a,q)=1$,
$$
\left\|\frac{a(m-n)}q\right\|\geq\frac1q,
$$
whereas $|\beta(m-n)|\leq1/(2q)$. Hence $\|\alpha m-\alpha n\|\geq1/(2q)$.
The points $\alpha m$ in one block are therefore $1/(2q)$-separated modulo one. Order them by distance from the nearest integer. Apart from a bounded number of endpoints, the $j$th closest point has distance $\gg j/q$, and consequently
$$
\sum_{m\text{ in one block}}\min(R,\|\alpha m\|^{-1})
\ll R+q\sum_{1\leq j\leq q}\frac1j
\ll R+q\log q.
$$
Multiplying by $O(M/q+1)$ gives
$$
\boxed{
\sum_{m_0\leq m<m_0+M}\min(R,\|\alpha m\|^{-1})
\ll(\log q)\left(\frac{MR}{q}+M+R+q\right).}
$$
This is the <reciprocal fractional-part sum near a rational> estimate.
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