I know about the improper integral issue, but analyzing the graph we see that $\tan (x)$ is odd around $\pi/4$, so if we say that the positive area is $+\infty$, we know that the other one is $-\infty$.
I know that infinities can't be summed, however we know that, because of odd condition, they are the "same infinity." So why isn't it zero?
$$\int\limits_0^{\pi/2}\tan(2x)\ dx = \int\limits_0^{\pi/4}\tan(2x)\ dx + \int\limits_{\pi/4}^{\pi/2}\tan(2x)\ dx $$ For the integral to exist, we need existence of both those integrals. Because those integrals do not exit, therefore neither does their sum.
There is, however, something called Cauchy's principal value, where we approach the limit at the same pace:
$$\text{p. v.}\int\limits_0^{\pi/2}\tan(2x)\ dx = \lim\limits_{\varepsilon\to 0^+}\left( \int\limits_0^{\pi/4-\varepsilon}\tan(2x)\ dx + \int\limits_{\pi/4+\varepsilon}^{\pi/2}\tan(2x)\ dx \right)= 0 $$
Indeed the value is 0 in this case.