Is it possible that $\sum n a_n^2$ converges but $\sum a_n$ diverges?

You could probably have success with $a_n=\frac1{(n+1)\log(n+1)}$.

$$\sum_{n\ge 2}\frac1{n\log(n)^2}$$

converges by twice applying the Cauchy condensation test or the integral test.

Let $$a_n=\frac1{n\ln n}$$ then using the integral test the series $$\sum_n a_n$$ is divergent but the series $$\sum_n na_n^2$$ is convergent.

An example such that $\sum\limits_nna_n^2$ converges and $\sum\limits_na_n$ diverges: $a_n=\frac1{n\log n}$.