Is there a sequence such that $\sum{a_n}$ diverges but $\sum{na_n}$converges?
No, there is no such sequence. Suppose that $\displaystyle \sum_{n=1}^\infty na_n$ converges. Then, since $\{\frac{1}{n}\}_{n\geq 1}$ is monotone and bounded (it goes to zero), we have $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n}na_n$ converging by Abel's test (or by Dirichlet's test); see Abel's test and Dirichlet's test .
The idea is (see applications of Abel transformation on Wikipedia):
If $\sum_n u_n$ is convergent and $v_n \to 0$ monotonically, then $\sum_n u_n v_n$ is convergent.
Here, take $u_n=na_n$ and $v_n=\frac 1 n$.
So the answer is no.