*Why* does normal reaction shift? For a block moving on a surface(due to some external force acting at the topmost point), if possible, the normal force shifts forward to maintain rotational equilibrium(net torque=0). Why does this happen? Why is rotational equilibrium preferred by nature?
 A: 
Why does this happen?

Because, otherwise the floor would break...
A normal force is nothing more than a "holding up" force. The floor doesn't care about balancing out forces or about balancing out torques. It just cares about not breaking.
When you push sideways at the top of the box, then not only does the weight pull down but also the torque tries to make the corner move downwards (and the opposite corner upwards). So this corner now pushes more downwards than before (and the opposite corner less than before).
There is a normal force at every contacting point, and the one at the corner now increases (while the one at the opposite corner decreases), and you have a gradual change in the normal force over the contacting surface. In many situations it is not easy to work with such a gradual spread of normal forces - so they are summed up into one normal force and averaged into acting at one point (in the same way that you sum up gravity in each particle and call it weight and then average that into pulling in just the centre of mass). This point will then shift leftwards, if there is more total normal force on the left side.

Why is rotational equilibrium preferred by nature?

Were there no floor, then the torque would not be balanced. And no rotational equilibrium would be achieved. So it is not accurate to say that "rotational equilibrium" is preferred by nature. Rotational equilibrium (as well as linear/translational equilibrium of course) just happens to be the case when things are not moving.
