Why is diboron (B2) paramagnetic?
For the same reason $\ce{O2}$ is paramagnetic: degeneracy.
The filled orbitals are $\sigma(\mathrm{1s})^2$, $\sigma^*(\mathrm{1s})^2$, $\sigma(\mathrm{2s})^2$, $\sigma^*(\mathrm{2s})^2$.  The half-filled orbitals are $\pi(\mathrm{2p}_x)^1$ and $\pi(\mathrm{2p}_y)^1$.  

But you're right to think that this is unexpected:  why are the $\pi(\mathrm{2p})$ orbitals  lower than the $\sigma(\mathrm{2p})$? That's the reverse of what we are told to expect! I've never encountered a rationalization that satisfied me. 

See Matt Law's lecture notes for the expected and observed ordering:
MO Diagrams for Diatomic Molecules; Chapter 5; Friday, October 9, 2015; PDF via the Internet Archive.