A few thoughts to help you on your way.
When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work / time = force * velocity ). If you are changing the velocity of an object, you are changing its kinetic energy: if it's slowing down, it gives energy back to you; if it's speeding up, you need to give it more energy.
If the elevator car is not moving, no WORK needs to be done. You still need a FORCE - but you could tie a knot in the cable and turn off the power: the elevator car will stay in place, without electricity, without heat being generated...
The helicopter example is different. The only reason a helicopter can hover is because it is pushing air down. Every second that it hovers, it needs to move a volume of air at a certain speed. In this case, the helpful equation is
$$F\Delta t = \Delta p$$
The change in momentum of the air tells you the force that you can get. This can be done by moving a large volume of air a little bit, or moving a little bit of air by a lot. Both situations will give you the same momentum, but since energy goes as velocity squared, the larger blades will be more efficient (up to the point where the drag of the blades becomes an important factor).
To solve the problem you stated, you need to know the size of the blades of the helicopter. Making some really simplified assumptions (there is at least a factor 2 missing in this - but just to get the idea): if you have a helicopter blade that sweeps an area $A$ and pushes air of density $\rho$ down with velocity $v$ then the force is
$$F = m \cdot v = (A\rho v)\cdot v = A\rho v^2$$
and the power needed (kinetic energy of the air pushed down per second) is
$$E = \frac12 m v^2 = \frac12 (A \rho v) v^2 = \frac12 A \rho v^3$$
If we assume $A=3m^2$ (roughly 1 m radius), and $\rho=1 kg/m^3$, then for a force of $500 N$ we need a velocity
$$v = \sqrt{\frac{F}{A\rho}} = \sqrt{\frac{500}{3}} = 13 m/s$$
and the power required is
$$E = \frac12 A \rho v^3 = \frac12\cdot 3 \cdot 13^3 = 3.3 kW$$
This is a bit higher than the 2kW you have available. The solution would be to increase the size of the rotors - the larger the area, the lower the velocity of the air, and the better off you are.
As for the pulley and string, or fixed string - see the comments I made about the elevator. When nothing moves, no work is done. In the case of the helicopter, although the helicopter doesn't move, the wings (blades of the rotor) do - and so does the air that is being moved (and whose motion provides the force needed to keep the helicopter in the air).
I hope that clears up your understanding.
Best Answer
First of all, the work needed depends on the way in which the final velocity is reached. If you write the expression for the work done by the drag force you obtain
$$W=-\int F ds = -b \int v^3 dt$$
and it's easy to see that if the object is carried to its final speed in a very short time $W$ can be arbitrarily small because $v$ in the integrand is always less than the final value, and the integration time can be reduced at will.
As a concrete example, let us suppose that the object is accelerated by an engine which has a constant power $P$. We can write
$$\frac{d}{dt} \left(\frac{1}{2} m v^2 \right)=P-bv^2$$
which can be explicitly integrated:
$$\frac{m}{2} \int_0^{v^2} \frac{dv^2}{P-bv^2} = t $$
which gives
$$\log \left( 1-\frac{b v^2}{P} \right) = -\frac{2bt}{m} $$
and
$$v^2 = \frac{P}{b}\left( 1-e^{-\frac{2bt}{m}} \right) $$
The maximum speed that can be reached is $\sqrt{P/b}$. Now, let us suppose for simplicity that $P$ is large compared with the power of the drag force. In this case we can expand the exponential at the first order obtaining
$$v^2 \simeq \frac{2Pt}{m} $$
The work done by the drag force can now be evaluated using the integral in the first equation,
$$W=-b\int \left( \frac{2Pt}{m} \right)^{3/2} dt =-\frac{4}{5}\sqrt{2} b t \left(\frac{P t}{m} \right)^{3/2}$$
which can be expressed as a function of the speed
$$W = -\frac{b m}{5 P} v^5$$
As expected from the general initial discussion, $W$ can be reduced by increasing the engine power $P$.