Let $(a_n)$ be a strictly increasing sequence of positive real numbers, and denote $\Delta a_n:= a_{n+1} - a_n.$
Suppose $\displaystyle\sum \frac{1}{a_n} $ diverges. Then does $\displaystyle\sum \frac{1}{n\Delta a_n}\ $ diverge?
My first thought for counter-examples was to try $a_n = n\log n.$ But I'm pretty sure $\displaystyle\sum \frac{1}{n\left( (n+1)\log(n+1) - n\log n \right)}\ $ diverges.
I can't think of a good proof strategy for the affirmative.
It suffices to show the following
Lemma 1. If $a_n>0$ and $\sum_{n=1}^\infty \frac1{na_n}<+\infty$, then $$\sum_{n=1}^\infty\frac1{a_1+a_2+\cdots+a_n}<+\infty.$$
Indeed, if $(a_n)$ is a strictly increasing positive sequence and $\sum_{n=1}^\infty\frac1{n\Delta a_n}<+\infty$, then we let $b_n=a_{n}-a_{n-1}>0$ for $n\geq2$ and $b_1=a_1>0$, and hence $\sum_{n=1}^\infty\frac1{nb_n}\leq \frac1{a_1}+\sum_{n=1}^\infty\frac1{n\Delta a_n}<+\infty$. By Lemma 1, we have $\sum_{n=1}^\infty\frac1{a_n}=\sum_{n=1}^\infty\frac{1}{b_1+\cdots+b_n}<+\infty$. This gives an affirmative answer to OP's problem.
Lemma 1 is equivalent to the following
Lemma 2. If $a_n>0$ and $\sum_{n=1}^\infty a_n<+\infty$, then $$\sum_{n=1}^\infty\frac{1}{\sum_{j=1}^n\frac1{ja_j}}<+\infty.$$
By Cauchy-Schwarz, we have $$k=\sum_{j=1}^k1=\sum_{j=1}^k\sqrt{ja_j}\cdot\sqrt{\frac1{ja_j}}\leq \sqrt{\sum_{j=1}^kja_j}\times \sqrt{\sum_{j=1}^k\frac1{ja_j}},$$ then $$\frac1{\sum_{j=1}^k\frac1{ja_j}}\leq \frac1{k^2}\sum_{j=1}^kja_j\leq2\left(\frac1k-\frac1{k+1}\right)\sum_{j=1}^kja_j.$$ Let $A_k=\frac1k\sum_{j=1}^{k-1}ja_j$ for $k\geq2$ and $A_1=0$, then $$\frac1{\sum_{j=1}^k\frac1{ja_j}}\leq2a_k+A_k-A_{k+1}.$$ Summing the above inequality for $1\leq k\leq n$ and using $A_n\geq 0$ for $n\geq 1$, we obtain $$\sum_{k=1}^n\frac1{\sum_{j=1}^k\frac1{ja_j}}\leq2\left(\sum_{k=1}^na_k\right)-A_{n+1}\leq 2\sum_{k=1}^na_k.$$ The proof of Lemma 2 is complete now.
