1) What you are saying seems correct to me that there is already the Normal force with the same magnitude as the gravitational force mg on the particle/mass you are lifting. BUT it is only right till the time you hold it. It is justified as long as you hold it(just hold it still in one place, and the mass is not moving). Now, since the acceleration is zero, the only forces acting on the mass you are holding are normal and gravitational.
<PERSON>'s second and third law satisfied (action-reaction pair and the net force is 0 since net acceleration while only holding is 0)
2) When you are trying to lift the mass, you need to accelerate it from its rest position in your hand, it has some inertia(inherent property of mass). 766
SO, we need to apply a force on it, which you termed "extra" apart from the normal reaction(which you denoted as F)
In all, the action-reaction pair cancel out each other whether you are holding your mass or accelerating it.
But, there is the force you apply to it to change its momentum. Let that force be F(which you said extra-to lift the mass-to accelerate it from rest resulting from action-reaction)
NOTE: You are applying a net force of N+Force(to accelerate from rest)>mg if F>0. So, there is a net force.
The equation will be: N+F-mg=ma(net)
N+F=m(g+a)
Since N=mg
F=ma(net)
So, your N component is resolved by the mg.
Also, in your equation,(which is wrong), N+F=mg, you are stating that if F is not equal to 0, then N is less than mg, but N and mg are action-reaction pairs.
It wont sink in at once-read it again, Good Day!
EDIT 1: See to make an object move at a constant velocity, it first needs to accelerate to that velocity in some earlier time, then the net force =0 can be done via the intervention of you.
EDIT 2: As <PERSON> pointed out, I would first like to say that I know what I had done. I did it just to make this thing simpler for the questioner. I have put it in the comments section that N and mg are not action-reaction pair. 766