Let $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$ be the measure space where $\mu$ is the counting measure. I know that, if $f$ is a nonnegative function, then $$\int_\mathbb{N} f \,d\mu = \sum_{n=1}^{\infty}f(n).$$
If $f$ takes any sign, then $f$ is integrable if and only if $\sum_{n=1}^\infty |f(n)|<\infty$, and in such a case $$\int_\mathbb{N}f\,d\mu=\sum_{n=0}^\infty f^+(n)-\sum_{n=0}^\infty f^-(n)=\sum_{n=1}^\infty f(n).$$
If I have a function $f$ such that $\sum_{n=1}^\infty |f(n)|=\infty$, is there any way to interpret $\sum_{n=1}^\infty f(n)$ as an integral?
Yes, you simply ask for $\int_0^R f \; d\mu$ to converge as $R\rightarrow \infty$. This is perfectly equivalent to asking for the series to be conditionally convergent.