Can a gravitational field be focused like a magnetic field? A pyramidal magnet will focus a greater magnetic field at the top of the pyramid than at the bottom.  See video here   Is there any evidence that the same holds for a gravitational field?  Perhaps the peak of a mountain will have a stronger gravitational field than would normally be the case for flat land.   Or perhaps the whole planet would need to be pyramidal shaped for this to occur.
 A: I used numerical integration to find the gravitational field at the top (i.e., apex point) and bottom (i.e, center of the square base) of an equilateral square pyramid. The force at the top is only 64.9% of that at the bottom.
Details:
The gravitational force on a test mass $m$ due to an arbitrary mass distribution with density $\rho(\mathbf r')$ can be calculated by considering the arbitrary mass distribution to be a collection of infinitesimal point masses $dM=\rho\, d^3\mathbf r'$. The infinitesimal force on $m$ at position $\mathbf r$ due to $dM$ at position $\mathbf r'$ is
$$d\mathbf F(\mathbf r)=-Gm\rho(\mathbf r')\,d^3\mathbf r'\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}.$$
(This is just the fancy version of
$$\mathbf F=-\frac{GMm}{r^2}\hat r$$
for when neither mass is at the origin and one of them is infinitesimal.)
The total force is found by doing an integral over the mass distribution to add up the infinite number of infinesimal contributions:
$$\mathbf F(\mathbf r)=-Gm\int_V\rho(\mathbf r')\,d^3\mathbf r'\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}.$$
For an object of uniform density, the $\rho$ is just a constant and comes out of the integral.
An equilateral square pyramid with side length $L$ has height $h=L/\sqrt2$. I took the pyramid to have its base vertices at $(\pm L/2,\pm L/2,0)$ and its apex at $(0,0,L/\sqrt2)$. At distance $z$ above the base, the pyramid extends between $\pm(\frac{L}{2}-\frac{z}{\sqrt2})$ in the $x$ and $y$ directions.
So the integral over the pyramid is
$$\mathbf F(x,y,z)=-Gm\rho\int_0^{L/\sqrt2}dz'\int_{-(\frac{L}{2}-\frac{z'}{\sqrt2})}^{\frac{L}{2}-\frac{z'}{\sqrt2}}dx'\int_{-(\frac{L}{2}-\frac{z'}{\sqrt2})}^{\frac{L}{2}-\frac{z'}{\sqrt2}}dy'\frac{(x-x')\hat x+(y-y')\hat y+(z-z')\hat z}{[(x-x')^2+(y-y')^2+(z-z')^2]^{3/2}}.$$
Taking $G$, $m$, $\rho$, and $L$ as 1, I used Mathematica's NIntegrate function (with the option method->"LocalAdaptive") to numerically integrate this at the top, $(0,0,1/\sqrt2)$ and the bottom $(0,0,0)$.
The results were
$$\mathbf F_\text{top}=(0,0,-0.961204)$$
and
$$\mathbf F_\text{bottom}=(0,0,1.48096).$$
A: Magnetic fields are fundamentally different to gravitational fields, even in the non-relativistic case, because there are no magnetic monopoles. By contrast, every piece of mass looks like a monopole.
The Maths
Using standard Newtonian gravity, the gravitational potential $U(\vec{x})$ of a mass distribution $\mu(\vec{y})$ is expressible via the Green's function for Poisson's equation:
$$V(x) = \int d^3 y \frac{-G \mu(y)}{|x-y|}$$
giving a gravitational field $\vec{g}(\vec{x}) = -\nabla V(\vec{x})$. Note that for the simple case of $\mu(y) = \delta^3(y-y_0)$, you recover the Newtonian gravity formula. Note that there is no antimass, $\mu \ge 0$.
This essentially says that the field from any mass distribution just looks like a point mass that's been 'smeared,' i.e. not that different to a sphere.
For a uniform density cone of mass M (nicer to parameterise than a square pyramid), the vector potential at a point in cylindrical coordinates $\rho, \phi, z$ is given by
$$\frac{-3MG}{\pi R^2 h}\int_0^R d\rho' \int_0^hdz' \int_0^{2\pi}d\phi' \left( (z-z')^2 + \rho^2 + \rho'^2 - 2\rho\rho' \cos(\phi')\right)^{-1/2}$$
This is not particularly illuminating, so it's more physically insightful to look into an approximation scheme.
Multipole Moments
You may wish to learn about the multipole expansion - a systematic way of expanding distributions of charge or mass, writing them as a sum of a monopole, a dipole, a quadrupole and higher moments. The potential from a monopole decays as $1/r$, from a dipole as $1/r^2$, from a quadrupole as $1/r^3$ and so on.
The monopole moment decays more slowly than all higher moments, so if it's nonzero it will eventually dominate when far away from the distribution. Since there's no such thing as anti-mass, monopole moments of mass distributions always win. However, the monopole moment of an arrangement of magnetic dipoles is always exactly zero.
Polarisability
The most critical difference between the magnetic pyramid and the mass pyramid is the fact that one is polarisable. Magnetic media are filled with spins that react when an external field is applied. In the special case of ferromagnets, nearest-neighbour spin-spin interactions collaborate to align with the applied field, strengthening it by a factor of 100 or more. The geometry and connectivity of the material will actually affect how the spins communicate, and hence the bulk polarisation. By contrast, mass just sits there being a weight - there is no material that acts as a field amplifier in the same way.
A: The answer is 'Yes'.  There is an exact analogy between classical magnetostatics and Newtonian gravity.  
For example, the magnetic field at the pointed top of a uniformally axially-magnetized pyramid or cone approaches infinity. The field can be viewed as arising from a shell of magnetic poles (charge) where the magnetization terminates on the surface of the cone.  Likewise the gravitational field at the point of a conical hollow shell of mass can also approach infinity. 
In magnetism, there are inevitably some balancing negative poles somewhere, but that is irrelevant since they can be moved arbitrarily far away with a suitable magnet design (e.g. a semi-infinite magnetized needle) 
There are some 'details' of course.  Infinite magnetic fields cannot be achieved in practice because there are no magnetic materials capable of sustaining infinite fields without demagnetizing and so the magnetic poles cannot be confined to an infinitesimally thin shell at the surface. Likewise matter cannot be confined to an infinitesimally thin shell and still retain any mass.  
A calculation of the gravitational field at the pointed top of a hollow pyramid of osmium (or, better still, neutron star stuff) will show an inordinately large gravitational field. 
