Why does it have to be so?

Think as to how you to have a gravitational force always equal and opposite to another type of force. Is it a situation which is possible?

@Farcher So you say that roots of normal force are gravitational ? Isn't it that normal reaction force has more to do with body lattice and it's strong,weak atomic forces ? What about centrifugal pseudo force, does it have same type as centripetal force too ?

@AgniusVasiliauskas Pseudo forces do not have Newton third law pairs. I have added to my answer in response to your other questions.

Again a down vote which I have no objection to bar no reason has been given as to why it was given.

Although these two forces have the same magnitude and are opposite in direction they are not a Newton third law pair for the following reasons. They act on the same body Are you sure ? I always thought that body weight acts on table and table normal force acts back onto body. No ?

@AgniusVasiliauskas I think you have misunderstood what I tried to write. Newton third law force pairs must act on different bodies whereas the two forces that I initially wrote about both act on the same body.

No, you misunderstood me. What I'm trying to say is that body weight and normal force acts on different bodies, not the same. Thus qualifies for a Newton force pair. Body weight acts on table(support) surface and table surface acts back with normal reaction force on body which lies on it.

We just draw weight vector starting from a body COM in free body force diagram, however it's not actually what is happening. Weight is a force which acts on support, thus technically weight vector should be drawn starting at body-table surface contact point(s)

@AgniusVasiliauskas "Weight is a force which acts on support, thus technically weight vector should be drawn starting at body-table surface contact point(s)" No, weight is the force of gravitational attraction on a body due to the Earth and for a extended body acts at the body's centre of mass.

Don't mess Newton gravitation law with weight - it's different things. Space shuttle in orbit "feels" gravitational force of Earth, however it still has no weight ! Because it has no "fixed" support to act upon which would not be falling too. Weight acts upon support surface and gravitational force acts upon body COM - they are different things. Gravitational force may or may NOT induce weight of body.

Besides, you mess a little bit with a reason of normal reaction force. Normal force reacts not to gravity, but to weight. It doesn't care about gravity. For example put body in container,- full of water-, at the bottom of it. Then container's bottom normal force will push body upwards with force of magnitude $F_{weight}-F_{buoyancy}$. Normal force doesn't care about gravity at all, just about "total pushing force against surface"

@Bill you miss the point. @Farcher's point is that Normal+Weight pair is NOT Newton force pair in third law, because he said that they targets SAME body. I'm saying, that they targets DIFFERENT bodies. As in your example too. Buoyant force targets rock surface(s), and rocks weight targets water column below it. Thus, it's perfect Newton third law force pair. So @Farher explanation about table-body system is WRONG.

These two forces can't act upon same body, simply because in such case there would be totally no reason for reactive normal force to arise ! Reactive force $F_{ba}$ is summoned only by $F_{ab}$. If weight and normal force acts exactly on same body, then one force can't be the reason of another. Which is not the case in here. I don't know how to explain better. Hope that helps.

The body has no net force on it and has two equal magnitude and opposite direction forces on it. The gravitational attraction of the Earth and the force due to the table. They are not a Newton third law pair. Unrelated the the original question you define weight one and I choose another way as explained in the Wikipedia article weight.

The contact force on the body due to the table is the Newton third law pair to the contact force on the table due to the body.

@AgniusVasiliauskas No, I don't miss the point. Farcher is correct that the 2 forces of a true N3L pair cannot target the same object. The normal force acting on the object and the force from the gravitational field interaction ARE acting on the same object and cannot be the N3L pair. They do NOT target different bodies if you do the analysis correctly.

@AgniusVasiliauskas Consider a system with no gravitational field consisting of a spaceship and an astronaut. If the astronaut pushes against the seat in the spaceship, there arises a pair of interaction forces, which we would designate as normal forces. They arise because of the electromagnetic forces holding the molecular structures of the astronaut/spacesuit and the seat together, preventing them from passing through each other. The normal force has NOTHING to do with a gravitational field and everything to do with the electromagnetic forces.

@Thomas Because forces are always interactions of 2 things, and they are symmetric, as far as we can tell, based on symmetry and isotropy of the universe. The fields which define the interactions do not seem to have a preferred direction in the universe, so every direction is equal. The strength of the interaction "forward" is the same as it is "backward." Symmetry makes the force directions directly opposite.

@BillN, Hey do not misinterpret me. Where I said that normal force has something to do with gravitational field ? Seems that you argue with yourself. I said that normal force is reactive force due to "total force pushing surface". That may or may not include weight.

The normal force acting on the object and the force from the gravitational field interaction ARE acting on the same object and cannot be the N3L pair Yes, I am not talking about gravity. I am talking about weight. And weight+normal force IS N3L pair, because acts on different bodies !

@Farcher I choose another way as explained in the Wikipedia article weight No you are not, because for you weight acts on same body COM as normal force do. So your definition is different than that in wikipedia. If you would be more careful you would notice there sentence "operational definition, the weight of an object is the force measured by the operation of weighing it, which is the force it exerts on its support"

@AgniusVasiliauskas All I can say is that I stand by my answer which did not have the word weight in it.