How much work is done to maintain vertical position (hovering) of a drone? Ignoring all losses in the efficiency of the means to lift the drone, this would reduce down to resolving the key elements of the equation involving a) the acceleration of gravity and 2) the time, and 3) the mass. So this is really just a math problem given that we assume the acceleration of gravity to be 9.807 m/s² and the time to be 60 seconds, and the mass to be 1 Kg.
An equivalent question would be to ask: "What work is done to accelerate a mass of 1 Kg along a flat horizontal plane at a rate of 9.807 m/s² for 60 seconds?" I am not interested in the kinetic energy because it does not change in the case of the hovering drone, only the work that is done over time.
 A: Your horizontally accelerating drone is not equivalent to the hovering drone. The fact that the hovering drone is not moving means that the energy expenditure can be made arbitrarily small.
In order to stay hovering, the drone needs an upward force of $F_{drone}$. The drone hovers by pushing air down, relying on Newton's third law to have the air push the drone up. In a small amount of time $\Delta t$ the drone's propellers push a mass of air $m_{air}$ down at a certain velocity $v_{air}$ (you can feel this air moving if you place your hand underneath a hovering drone). The force felt by the air (and thus by the drone in the opposite direction) is
$$F_{air} = m_{air}a_{air} = m_{air}\frac{v_{air}}{\Delta t}$$
Or, rearranging
$$F_{air}\Delta t = m_{air}v_{air}$$
Using the third law,
$$F_{drone}\Delta t = m_{air}v_{air}$$
So, in order for a drone to hover for a certain time period ($\Delta t$), it needs to push down a certain amount of air ($m_{air}$) at a certain velocity ($v_{air}$). Notice that the drone designer can choose to push down more air at a lower velocity or less air at a higher velocity.
Now, how much power does this require? The air gains kinetic energy by the action of the propellers. In the absence of any other inefficiencies, the drone engine must supply enough power to give kinetic energy to the air. The kinetic energy that must be supplied by the drone engine in time $\Delta t$ is given by
$$K_{air} = \frac 1 2 m_{air} v_{air}^2$$
Notice that velocity is squared but mass is not. If I double the mass and halve the velocity, I get the same force with half the power. More generally, if I multiply the mass of air by $k$ and divide the velocity by $k$, then the power needed to produce the required force is reduced to $1/k$. It is more efficient to push a lot of air down slowly than to push a little air down quickly. In the ideal case, with propellers that very long and very low-density, the power needed can be reduced to arbitrarily small amounts.
It is the details of the mechanism of hovering that determine how much power is needed. For example, here's a video of a human-powered helicopter. Notice the large, slow-moving propeller blades that are designed for efficiency.
To take this to a somewhat ridiculous extreme the drone can hover with the engines off by placing the drone on a table at the desired height.
A: The drone must push the air (of density $\rho$) down, from beneath the rotor blades covering area $A$ - at a speed $v$, so that the change of momentum of the mass of air per second equals the weight of the drone, of mass $m$.
(this is the impulse equation, Force equals the rate of change of momentum.  From Newton's third law the upward force on the drone is equal to the downward force on the air).
The volume of air pushed down per second is $Av$ and the change in momentum of this air is $Av\rho \times v$
i.e. $$mg = Av^2\rho$$
and $$v^2 = \frac{mg}{A \rho}$$
There will be a minimum energy, per second, needed to maintain the drone stationary, equal to the kinetic energy of the air that's pushed down.
$$K.E = 0.5(Av\rho)v^2 = 0.5Av^3\rho $$
so
$$ K.E = 0.5\frac{(mg)^{3/2}}{(A \rho)^{1/2}}\tag 1$$
then we'd multiply by 60 seconds.
Taking $m=1$ $g=9.8$ $\rho = 1.3$
gives an approximate answer for the energy needed of  $$E = \frac{807}{\sqrt{A}}\tag2$$ in Joules, for the 60 seconds.  It depends on the area, taking  $A=0.1$ square meters, for example, gives about $$E=2550J$$
A: If you ignore all losses due to inefficiency then the work for hovering is 0.
A drone is horribly inefficient. To make things hover efficiently you can use a superconductor and a magnetic field which requires 0 energy. There are also gyro-stabilized magnets that waste very little energy hovering.
Here is a picture of a disk hovering with 0 energy

