Is the language $L = \{ a^ib^j \mid i\ \nmid\ j \ \} $ context free?
If I'm not mistaken, you can pump $\bar L$ using $\sigma = a^{n}b^{n^{2}}$, because $n \mod n^{2} = 0$. The result is that $\bar L$ is not context free. The property that you mentioned has an "iff", then $L$ is not context free.