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For every $\varepsilon>0$, only finitely many $q>0$ satisfy

$$\sum_{\sigma(d)/d=q}\frac1d\ge\varepsilon.$$

The equality case at mass one is already proved in the fibre theorem.

Proof. Use its representation $S_q=\sum_\gamma1/(\gamma s_{\gamma,q})$. The convergent sum $\sum_\gamma1/\gamma$ has a tail less than $\varepsilon/2$ outside some finite set $F$ of powerful numbers. This tail bound is uniform in $q$, because $s_{\gamma,q}\ge1$ when finite. If $S_q\ge\varepsilon$, then

$$\sum_{\gamma\in F}\frac1{\gamma s_{\gamma,q}}\ge\varepsilon/2.$$

At least one of its $|F|$ terms is at least $\varepsilon/(2|F|)$. Therefore, for some $\gamma\in F$, the finite positive integer $s_{\gamma,q}$ is at most $2|F|/(\varepsilon\gamma)\le2|F|/\varepsilon$. There are only finitely many such pairs $(\gamma,s)$, each determining $q=\sigma(\gamma s)/(\gamma s)$. This proves finiteness. $\square$

Source. Tao, published paper, published p.818, Remark 4.6. This page uses that published version.

Bears on. Problem 49.