One way to show this is meromorphic is via Morera's theorem, which is as follows.
Suppose that for every simple closed curve $\gamma$ in some domain, such that $\gamma$ does not wind around any point in $\mathbb{C}$ that is not in the domain, $\displaystyle\int_\gamma f(z) \, dz=0$. Then $f$ is holomorphic in that domain.
Now look at $ \int_\gamma f(z)\,dz = \int_\gamma \sum_{n\in\mathbb{Z}} \frac{dz}{(z-n)^2}. $
Either Fubini's theorem or Tonelli's theorem implies that the last expression above is equal to $ \sum_{n\in\mathbb{Z}} \int_\gamma \frac{dz}{(z-n)^2}. $ (Fubini's theorem implies the order of integration can be reversed in iterated integrals in which the integral of the absolute value is finite. The sum is an instance of a Lebesgue integral with respect to counting measure. Tonelli's theorem gets the same conclusion in case the function being integrated is everywhere non-negative, regardless of whether its value is finite or not.) The last integral above is $0$ since $\gamma$ doesn't wind around $n$. Hence the conclusion of Morera's theorem holds.
This shows that $f$ is holomorphic in $\mathbb{C}\setminus\mathbb{Z}$, and hence meromorphic if it has a pole at each point in $\mathbb{Z}$.
For any $n_0\in\mathbb{Z}$, we have $ f(z) = \frac{1}{(z-n_0)^2} + \sum_{\begin{smallmatrix}n\in\mathbb{Z}\\ n\ne n_0\end{smallmatrix}} \frac{1}{(z-n)^2}. $
The second term can be shown to be holomorphic in $\mathbb{Z}\setminus\{n_0\}$ by the method used above. The first term has a pole of order $2$ at $n_0$.