The Enigma of Universal Gravitation Forces This is taken from a book called "Physical Paradoxes and Sophisms" by V. N. Lange.

1.22. The Enigma of Universal Gravitation Forces
  The law of gravitation can be written $F=\gamma\frac{m_1m_2}{R^2}$.
  By analyzing this relationship we can easily arrive at some interesting conclusions: as the distance between the bodies tends to zero the force of their mutual attraction must rise without limit to infinity.Why then can we lift up, without much effort, one body from the surface of another body (e.g., a stone from the Earth) or stand up after sitting on a chair?

 A: There is no paradoxon.
In your examples the distance $R$ between the bodies
(more exactly: the distance between their centers of mass)
is not zero.


*

*The distance between a stone and the earth is around 6400 km, but not 0.

*The distance between your body and the chair is around 0.5 m, but not 0.


Therefore the force $F=\gamma\frac{m_1 m_2}{R^2}$ is not infinite.
A: It is just a math exercise about limits: 
$F = \frac{GMm}{R^2}$. 
If $\rho$ is the density of the body with mass $M$ (supposed spherical for simplicity):  
$F = \frac{G\rho m 4\pi R^3}{3R^2}$.
So: $F = \frac{G\rho m 4\pi R}{3}$ goes to zero when $R$ goes to zero!
A: Consider the gravitational acceleration $g$ at a surface point $p$ of a massive compact body $M$ whose density $\rho$ is assumed to be bounded from above. Then $g$ is finite rather than infinite.


*

*If $M$ is a spherically symmetric ball, this is true due to Newton's shell theorem. The effective distance $r>0$ in Newton's formula then becomes the radius of $M$.

*Even without spherically symmetric this is true. At the surface point $p$, one can sum up hemispheres with center $p$ and radius $r$. The $1/r^2$-singularity in Newton's formula gets cancelled by a $r^2$-area-dependence of the hemisphere with thickness $dr$, so that $g$ is not infinite but grows (at most) linearly with the size of $M$. A similar estimate can be found from Gauss' law for Newtonian gravity.
If you like OP's question, you may also enjoy reading this Phys.SE post.
A: The reason for this is that the formula
$$F_G = G\frac{m_1 m_2}{r^2}$$
technically only gives the gravitational force between two objects that are of point size, i.e. that have size zero. When $r = 0$, that is saying that you have two infinitely concentrated masses infinitely close together. Does this sound like a good description of the situation of an object resting on the surface of the Earth? Is the Earth an infinitely concentrated mass (i.e. a singularity)?
Hence, you can only unqualifiedly apply this formula when that we are talking of objects so far apart that their size is negligible and they are "approximately pointlike", such as a spacecraft far from the Earth, or the planets in the Solar System (e.g. the distance between the Earth and the Sun is on the order of 150 Gm, but the Sun itself is only about 1 Gm in size and the Earth, 0.012 Gm, so both can be treated as roughly pointlike).
To treat extended bodies, we must use a continuum-mechanical description. Instead of the above formula, we must use the more general gravitational field, $\mathbf{g}$, which has a vector value at each point in space, and we require that each infinitesimal part of the extended body act individually as an infinitesimal source of gravitational field:
$$\nabla \cdot \mathbf{g} = -4\pi G\rho$$
where $\rho$ is the mass density at a given point in space, and the left hand term is the field divergence, which basically describes how much locally each point acts as a source of field. Moreover, the gravitational field should be conservative:
$$\nabla \times \mathbf{g} = \mathbf{0}$$
. And then we can integrate these to obtian $\mathbf{g}$, and then the force $\mathbf{F}_G$ on an object by $\mathbf{F}_G = m\mathbf{g}$.
Despite that, however, there is a "saving grace" that allows one to more simply use the simple first formula reasonably well to treat the situation of the object on the Earth. Using the more proper, extended-body description of gravitation we just gave, you can prove that for a spherically symmetric mass distribution $m_1$ of radius $R$, that the gravitational force on a point-like object of mass $m_2$ at a height $h$ above its surface is
$$F_G = G\frac{m_1m_2}{(R + h)^2}$$
and to the extent you can consider Earth a spherical object, this formula says that in effect, you can treat the Earth as being equivalent to a point-like gravitator with all its mass concentrated at its center, and thus the $r$ going into the formula for the pebble-on-ground scenario should be the entire Earth's radius, i.e. the distance from its center to its surface, and not simply the contact distance between the surfaces of the two objects. This is about 6371 km, or $6.371 \times 10^6\ \mathrm{m}$, with units at the scale of a human being (1-2 m), or $6.371 \times 10^9\ \mathrm{mm}$, with units at the scale of a pebble (1-10 mm). So clearly, your gravitational force will be quite a bit less than infinity once you use that large number instead of zero :)
A: Because in general gravitational attraction is much weaker than electromagnetic repulsion.
For example when you compare gravitational and electrostatic forces between two electrons, both forces are proportional to $R^{-2}$, but if you compare their actual values, you get that gravitation is around $10^{37}$ times weaker. 
A: Answers which merely refer to the distance between the centre of mass of one thing and the centre of mass of another are missing the point being made by Lange. 
When we sit on a chair, the distance between part of your body and part of the chair does tend to zero, so what happens to the force? If our atoms had point-like particles in them with finite mass, then, even though the masses are small, when those particles met there would be infinite forces. It is quantum theory that prevents these infinities, by showing how the mass is not in the end located in point-like regions of space. In classical physics one can get to the same conclusion by postulating that the density has to be finite everywhere.
(This answer amounts to a further comment on/exposition of the one by Qmechanic.)
A: We have the power to overcome this near infinite force of gravity because the electromagnetic forces generated by our muscles is much much stronger.
Every cell in our body burns on the order of 1 to 10 million molecules of ATP every second.
That’s when those cells are at rest.
During periods of high intensity, that ATP burning increases 1000 fold in muscle cells.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3005844/
A: The distance between the stone and the earth isn't zero (and therefore gravity force isn't infinite), but rather separated by electrostatic repulsion of the electrons, a distance on the order of 10^-10 m.  That is the equilibrium balance point between gravity attraction and electrostatic repulsion.
