I was computing the Fourier coefficient of the function: $$ F(t)=\left\{ \begin{array}{rl} F_0,&0<t<\pi,\\ -F_0,&\pi<t<2\pi, \end{array} \right. $$ with $F(t+2\pi)=F(t)$. Since this function is odd, only the $b_n$ coefficients do not vanish. However they are computed to be $$b_{2m+1}=\frac{4F_0}{(2m+1)\pi},\quad b_{2m}=0,$$ i.e. only the odd coefficients (among the $b_n$) survived. Is there any deeper reason for that (maybe some symmetry of F(t)) or it is just a coincidence?
Note: The trigonometric expansion I am using is $$F(t)=\frac{a_0}{2}+\sum_{n=1}^\infty\left[a_n\cos(nt)+b_n\sin(nt)\right],$$ so that $$b_n=\frac 1\pi \int_{-\pi}^\pi F(t)\sin(nt)dt.$$
The $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} F(t) \cos nt \, dt$ are zero because $F(t)$ is, as you have noted, odd. It also follows that the $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} F(t) \sin nt \, dt$ are non-zero.
Making use that $F(t)$ is odd about $t = 0$ means that you can write the coefficients $b_n$ as an integral 0ver $0 \to \pi$, namely $b_n = \frac{2}{\pi} \int_{0}^{\pi} F(t) \sin nt \, dt$ and since $F(t)$ is constant over this region gives $b_n = \frac{2 F_0}{\pi} \int_{0}^{\pi} \sin nt \, dt$.
Looking at this you can see that the $b_{2m}$ will vanish and the $b_{2m+1}$ will survive because over the range $0 \to \pi$, $\sin 2mt$ is odd about $t = \frac{\pi}{2}$ while $\sin (2m + 1)t$ is even (to see this plot a few examples). So the reason the even terms vanish and the odd terms survive can be related to the form of $F(t)$ being symmetric about $\pi/2$ (this is true for any $F(t)$ symmetric about $\pi/2$ not just $F(t) = F_0$).