Fourier Transform of Alternating Periodic Rectangular Pulse
The answer is simple.
I will give 3 points to solve it:


The Fourier transform is linear. Hence $ \mathcal{ F } \left\{ \alpha f \left( x \right) + \beta g \left( x \right) \right\} = \alpha \mathcal{ F } \left\{ f \left( x \right) \right\} + \beta \mathcal{ F } \left\{ g \left( x \right) \right\} $.
Shift in time $ f \left( x - {x}_{0} \right) $ equals multiplication by $ {e}^{-j \omega {x}_{0}} $ in Fourier domain.  
Instead of solving for the case above, think of the case you have 2 rectangular signals with twice the period with one multiplied by $ -1 $ and shifted. 


In your case, just have a look on the signal of the Positive Pulses. It has a period of $ 4T $.
You have another signal. The signal of the negative pulses.
It is basically the same signal as the positive one (It also has a period of $ 4T $) yet it is shifted by $ 2T $ and as it is multiplied by $ -1 $.
So if you know the transform of the positive one, follow my above points and you have the transform of the negative one and their sum.