There is a minimum amount of work to empty half the aquarium. I'll do that. I need to presume that we are lifting the water out to a height no greater than the top edge of the aquarium, which I will define as our zero point for the gravitational potential energy. Then the work done is equal to the change in potential energy of the water pumped out. (There's a missing negative sign here: I'm defining downward to be positive for simplicity, so we lose a negative sign.)
Segment the aquarium into thin planes of water, each at a depth x below the top of the aquarium, each plane of water having a height of dx. We wish to pump out all planes down to x = 1/2 m.
So consider the plane at a depth x. How much work is needed to pump this plane of water out of the tank?
dW = (Delta)PE = gx dm
where g = 9.8 m/s^2, and dm is the mass of the plane of water at depth x. And we know:
dm = (rho)dV = (rho)*L*W*dx
where (rho) is the density of water, and L and W are the length and width of the tank, respectively.
dW = (gx)((rho)LW dx) = ((rho)gLW)(x dx)
W = ((rho)gLW)*Int[x dx, 0 , 1/2] = ((rho)gWL/2)*(1/2)^2 - 0 = ((rho)gWL/8) = 2450 J