You cannot have a total vorticity with periodic boundary conditions, since if you take a path around all of your vortices, it will have a non-zero circulation. But you have periodic bc, so you can continuously deform that path to a point, and a point has zero circulation.
Mirror images are not quite the same as in electrostatics. We want periodic boundary conditions. To get periodic boundary conditions you imagine tiling the plane with your system. This will trivially be periodic and so you can just take a tile, and work with that. So you want the mirror image of a vortex to be a vortex.
(In electrostatics you usually use mirror images to enforce not periodic boundary conditions, but to enforce constant voltage. That's why you flip the sign of the mirror charge. Here we want periodic b.c. If you flipped the sign of the vortices I believe you would get *anti*periodic boundary conditions, but don't quote me on that.)
This evolving in imaginary time is presumably an "annealing" type of operation. You are free to run the GP equation on any initial condition you want. However, to cleanly see the interaction of the vortices, we want to the vortices to be in their ground state. Otherwise when we turn on the time evolution they will get rid of their excess energy by shedding waves and other junk.
One way to get to the ground state is to evolve you equation in "imaginary time". Your usual time evolution is $\exp(i\hat{H}t)$. If plug in $t = i\tau$ you get $\exp(-\hat{H}\tau)$. Applying this to a state exponentially suppresses the higher-energy components, so you get rid of the high energy stuff. This is related to finite-temperature (just replace $\tau$ with $\beta$ and you have the partition function), but for your purposes you can just consider it a convenient mathematical trick.
Note that since you're annealing anyway, the specific details of what state you start might not be so important, since you will end in the same place anyway (hopefully).
Finally, they are a little thin on the details, so if you plan to use this work, you should just send the authors an email asking for details.
Best Answer
In the case of electrostatics in free space, Laplace's equation holds whenever the domain in question is charge-free. This follows since $$\Delta \phi=\nabla\cdot\nabla\phi=-\nabla\cdot\mathbf{E}=-\frac{\rho}{\epsilon_0}=0$$ whenever $\rho=0$.
Not necessarily, the boundary conditions can be anything, not just constant.
In fact, if the boundaries are all at the same constant potential, then it's rather boring; the solution just becomes $\phi=\phi_b$ everywhere inside the domain, where $\phi_b$ is the constant boundary potential. In that case, $\mathbf{E}=-\nabla\phi_b=0$, which is why the inside of a hollow metal object has zero field (Faraday cage effect).