Pushing air downwards using a fan, to hover in air I believe that I am simply missing something over here, but cannot find it.
I want to just think theoretically, that if I use a fan (drone manner) to push air downwards, like a rocket, I would be pushed upwards.
So I need to apply a constant force of  upwards. In my case, it's 700N approx.
Now, much electrical power would be needed for the job?
 A: If you use a single fan, it will not only accelerate air downwards but also add a bit of swirl. Just as the downward acceleration results in an upward force (the desired lift), the rotational acceleration will result in a torque on the fan and motor assembly, so it will start to rotate in the opposite direction of the fan's rotation.
Either you use an even number of fans and create counter-rotating pairs of them, or you add some device to produce counter-torque. In most helicopters of the Sikorsky type this is a small vertical rotor at the end of a tailboom, or it is a jet of air (Google for NOTAR to learn more).
Once this is solved, start with the static thrust equation of a propeller:
$$T_0 = \frac{P\cdot\eta_{Prop}\cdot\eta_{el}}{\sqrt{\frac{2\cdot T_0}{\pi\cdot d_P^2\cdot\rho}}} = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{Prop}^2\cdot\eta_{el}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$
Nomenclature:
$\kern4mm T_0\kern7mm$Thrust at zero forward speed
$\kern4mm P\kern8mm$Motor power
$\kern4mm \eta_{Prop}\kern2mm$propeller efficiency. Could be anything between 0.5 and 0.85
$\kern4mm \eta_{el}\kern6mm$electric efficiency of motor and its controller
$\kern4mm d_P\kern6mm$propeller diameter
$\kern4mm \rho\kern9mm$air density
$\kern4mm \pi\kern9mm$3.14159265…
A: The required power is
$$W=\sqrt{\frac{\left ( mg \right )^{3}}{2pS}}$$
-S being the surface of the disc enclosing the fan
-g being gravity acceletation
-p is density of air (kg/m3)
-m is the mass being lifted
The demonstration is Rankine-Froude theorem, which relies on Bernouilli's principle in the air column of the fan to show that
$$T=2psv^{2}$$
-T being the lifting force
-v being the induced speed of air at the fan
From there you get in stationnary flight that gravity=lift:
$$mg=2psv^{2}$$
or
$$v=\sqrt{\frac{mg}{2pS}}$$
Then by multiplying the speed with the force(=mg) you get the power W in the first equation.
In your specific case you can replace "mg" by 700 N
