If $r_i=r_{i+t}$ for some $t$ and all $i$ you can split the sum into a finite sum and the rest $$\sum_{i=1}^t \frac{r_i}{i}+\sum_{i=t}^\infty \frac{r_i}{i}=\sum_{i=1}^t \frac{r_i}{i}+r_t\sum_{i=t}^\infty \frac{1}{i} $$ by your assumption. So the latter is just the harmonic series and your series diverges unless $r_t=0$.

This reduces your question to a finite question. Now the answer depends on what you call non-trivial. Eventually all $r_i$ have to be 0, but but of course solutions exist. For example $r_i=0$ for all $i \neq 1,2$ and set $r_1=1$ and $r_2=-2$.

If $r_i=r_{i+t}$ for some $t$ and all $i$ you can split the sum into a finite sum and the rest $$\sum_{i=1}^t \frac{r_i}{i}+\sum_{i=t}^\infty \frac{r_i}{i}=\sum_{i=1}^t \frac{r_i}{i}+r_t\sum_{i=t}^\infty \frac{1}{i} $$ by your assumption. So the latter is just the harmonic series and your series diverges unless $r_t=0$.

This reduces your question to a finite question. Now the answer depends on what you call non-trivial. Eventually all $r_i$ have to be 0, but but of course solutions exist. For example $r_i=0$ for all $i \neq 1,2$ and set $r_1=1$ and $r_2=-2$.

EDIT: Ok, I misread the question. And my answer is not correct. At least using the arguments above you can show that the series can only be conditionally convergent. Since otherwise, you could rearrange the series as follows

$$ r_1 \sum_{i \equiv 1 \mod t}\frac{1}{i}+\cdots+r_{t-1} \sum_{i \equiv t-1 \mod t}\frac{1}{i}$$. And these are essentially the harmonic series.