It's a good question, and it has a beautiful answer.

It is true that for any finite sum (referring to your first expression), one cannot have a complex pole. So the question is: how can the complex pole appear in the $\boldsymbol{n \to \infty}$ limit? (Spoiler: it can.)

The short answer is that in the $n\to \infty$ limit, the Green's function can develop a branch cut. What this means is that the function is more naturally defined on a larger Riemann surface, which means one can use this branch cut as a 'portal' to a new sheet. The complex pole lives on this new sheet! Not even M. Night Shyamalan could've thought of this one.

Let me illustrate this with an example. Suppose we have $$G(z) = \sum_n \frac{c_n}{z-\varepsilon_n}\qquad \textrm{with } c_n = \frac{1}{\pi} \frac{\Gamma}{\varepsilon_n^2 + \Gamma^2},$$ where we can think of the $\varepsilon_n$ being some equally spaced list of energies along the real axis. This function clearly has no complex poles. However, in the continuum limit, we can easily calculate (e.g. by using the residue theorem) that we obtain $$ G(z) = \int_{-\infty}^\infty \frac{1}{\pi}\; \frac{\Gamma}{\varepsilon^2 + \Gamma^2} \; \frac{1}{z-\varepsilon} \mathrm d\varepsilon = \left\{ \begin{array}{ccl} \frac{1}{z+i\Gamma} && \textrm{if Im } z > 0, \\ \frac{1}{z-i\Gamma} && \textrm{if Im } z < 0. \end{array} \right.$$ Clearly, this function still has no complex poles on the complex plane! However, we now have a branch cut on the real axis. In such a case, the function $G(z)$ is more naturally defined on a more general Riemann surface. In this very simple case, the situation is a bit funny: there are two complex continuations, depending on whether we come from the bottom or the top. In other words, $G(z)$ can be more naturally thought of as being defined on two disjoint Riemann spheres, being the function $\frac{1}{z+i\Gamma}$ on the other, and being $\frac{1}{z-i\Gamma}$ on the other. Clearly, on this more complete domain, we have complex poles! (Remark: these two separate functions correspond to the retarded and advanced Green's functions.)

It's a good question, and it has a beautiful answer.

It is true that for any finite sum (referring to your first expression), one cannot have a complex pole. So the question is: how can the complex pole appear in the $\boldsymbol{n \to \infty}$ limit? (Spoiler: it can.)

The short answer is that in the $n\to \infty$ limit, the Green's function can develop a branch cut. What this means is that the function is more naturally defined on a larger Riemann surface, which means one can use this branch cut as a 'portal' to a new sheet. The complex pole lives on this new sheet! Not even M. Night Shyamalan could've thought of this one.

Let me illustrate this with an example. Suppose we have $$G(z) = \sum_n \frac{c_n}{z-\varepsilon_n}\qquad \textrm{with } c_n = \frac{1}{\pi} \frac{\Gamma}{\varepsilon_n^2 + \Gamma^2},$$ where we can think of the $\varepsilon_n$ being some equally spaced list of energies along the real axis. This function clearly has no complex poles. However, in the continuum limit, we can easily calculate (e.g. by using the residue theorem) that we obtain $$ G(z) = \int_{-\infty}^\infty \frac{1}{\pi}\; \frac{\Gamma}{\varepsilon^2 + \Gamma^2} \; \frac{1}{z-\varepsilon} \mathrm d\varepsilon = \left\{ \begin{array}{ccl} \frac{1}{z+i\Gamma} && \textrm{if Im } z > 0, \\ \frac{1}{z-i\Gamma} && \textrm{if Im } z < 0. \end{array} \right.$$ Clearly, this function still has no complex poles on the complex plane! However, we now have a branch cut on the real axis. In such a case, the function $G(z)$ is more naturally defined on a more general Riemann surface. In this very simple case, the situation is a bit funny: there are two complex continuations, depending on whether we come from the bottom or the top. In other words, $G(z)$ can be more naturally thought of as being defined on two separate complex planes: on one we have $\frac{1}{z+i\Gamma}$, and on the other $\frac{1}{z-i\Gamma}$ on the other. Clearly, on this more complete domain, the functions are analytic (i.e. no more branch cut) and we have complex poles at $z = \pm i \Gamma$! (Remark: these two separate functions correspond to the retarded and advanced Green's functions.)

Ending up with a Riemann surface that is the union of two disjoint spaces is a bit unusual and is an artifact of $G(z)$ having a branch cut that cut the plane in two. In more general situations (where the branch cut doesn't extend over the whole real axis), the Riemann surface will be connected. It will still be true that to get to the complex pole, you'd need to cross the branch cut, into the new branch. (Since this new branch cannot be 'accessed' without a branch cut, it makes sense that there is no complex pole for any finite sum.)