Man in elevator, holding it, on a scale 
This is the scenario where my mass is $60 kg$, the mass of the elevator is $30kg$, and due to a malfunction, I have to hold myself and the elevator at rest. The question is, if there is a weighing scale under me, (of negligible mass), then what force will it measure (in newtons)?
My approach:
The total mass of the system is $90kg$ which is approximately $900 N$ for equilibrium, tension in the rope is $900N$ because I am giving a downward force of $900N$, so according to Newton's 3rd Law, upward force on me by the rope must be $900N$. This means I should go up, and the weighing scale must not show any reading. But that is not among the options to choose from.  Guidance rather than a direct answer would be appreciated.
 A: 


Now these two equations can be solved to get the $N$ , by eliminating $T$ . Remember, only the normal force is reading of the elevator.
A: The total force down is 900N. Thus, the total force up must be 900N. For the rope to be in equilibrium, the force up on each side of the pulley is 450N.
Your force down is 600N, the elevator's force down is 300N. If we carry through the math, that means your net force is 150N down and the elevator's net force is 150N up.
Assuming the weigh scale is considered part of the elevator, it would show 300N. In other words, your mass minus the elevator's mass.
Edit
Based on the options provided, I would say you should select the 150N option. However, I believe that is wrong. If we symbolize the masses instead of use numbers, then that answer implies the following:
The force due to gravity of you and the elevator is $-m_1g$ and $-m_2g$ respectively. Thus, in equilibrium, the tension on each side of the pulley would be $g(m_1+m_2)\over2$. And the reading on the scale then is $g(m_1-m_2)\over2$, or the normal force on you as 007 put it.
I believe that this is wrong because in the limit where the mass of the elevator, $m_2$ goes to 0, then it would suggest that the scale reads $gm_1\over2$, or half your mass when, in that limit, it should read your full mass.
What I believe is left out is that due to the tension from the rope, the elevator has a net force upwards and, when combined with your net force downward, the relative normal force the scale would need to provide is a sum of the two, in other words it is $g(m_1-m_2)$, your mass minus the mass of the elevator.
A: we can actually use $300 - T = 0$ (1) for equilibrium of the lift and $600 -T = 0$ (2) for equilibrium of the man as both systems are at rest.
Adding (1) and (2)
$900- 2T = 0$
$T = 450 N$
Taking into account the system of the man, the man exerts a force downwards and tension pulls him upwards.
Thus total force downward in the man's system is $600 N - 450 N = 150 N$
The reading of the weighing machine is $150 N$
