Can massless particle have effective mass? The effective potential was probably very familiar in many concepts. 
However, what about effective mass?
Suppose a massless particle. For simplicity, suppose it's not some superficial particle, i.e. it has  observable effect. 
Is it possible for such massless particle to gain an "effective mass" through dynamical interaction? 
For example, a photon could well obtain a $e^-\sim e^+$ pair in space, but I'm not sure weather it's a meaningful case. 
Further, what does it mean to four momentum for such effective mass, if it exists.  
 A: As has been mentioned in the comments a few times, the Higgs mechanism may be what you're looking for. Here's an example:
Suppose we have the following Lagrangian density:
$$\mathcal{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} + |D_\mu \phi|^2 -\mu^2 |\phi|^2 + \frac{\lambda}{2} (|\phi|^2)^2$$
Here $\phi(x)$ is a complex scalar field with magnitude squared $|\phi|^2$, $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is the electromagnetic field tensor, and $D_\mu \phi = \partial_\mu \phi - ieA_\mu \phi$ is the covariant derivative of the scalar field $\phi$. We can see that, as usual the photon is massless (i.e. there is no term proportional to $A^2 = A_\mu A^\mu$). However, if we consider the classical equation of motion for $\phi$ and look for constant configurations where $\phi = \bar\phi$, we get
$$-\mu^2\bar\phi + \lambda \bar\phi |\bar\phi|^2  = 0$$
and a similar equation for the complex conjugate $\phi^*(x)$. 
If $\mu^2 >0$ then the solutions to this equation of motion are $\bar\phi = 0$ and $\bar\phi = \sqrt{\frac{\mu^2}{\lambda}} \equiv v$.  Suppose that $\phi$ takes the non-zero value $v$ classically. Then it would be natural to expand the field around that value by writing$$\phi(x) = v + h(x)$$ for some new shifted field $h(x)$. You should check for yourself that if you plug this back into the original Lagrangian, you end up with a term which is proportional to $v^2 A_\mu(x) A^\mu(x)$ (assume that $v$ and $h$ are real valued). This is a mass term, and so the photon becomes massive and furthermore the value of its mass depends on the classical value of the scalar field $\phi$.
If you'd like a reference to check out on the Higgs mechanism and spontaneous symmetry breaking, chapter 11 of Peskin and Schroeder might be a good place to start. Also check out part (a) of the "final project" after chapter 13. 
A: The answer is no, a particle that has zero mass, as the photon, is a relativistic particle and the algebra of the Lorentz transformations does not allow it because of  energy and momentum conservation laws. The mass is the length of the energy-momentum  fourvector.

For example, a photon could well obtain a $e^-\sim e^+$ pair in space, but I'm not sure weather it's a meaningful case. 

No it is not, again because of energy momentum conservation laws.
The electrons if on shell have a mass each of about o.5MeV. The invariant mass of the the produced pair would have to be at least ~1MeV, whereas the photon has 0 mass. This means that the energy momentum of the system, will have to be different from the incoming, leading to violation of energy and momentum conservation. Pair production of photons has to happen with  interactions with another field 

In this instance the field of a nucleus Z.
A: A photon, when propagating in a photonic crystal, will have a nonzero effective mass in all bands, except possibly the point $\vec k=0$ in the lowest band. This effective mass has all the weird properties of that of electron in crystal: anisotropy, varying value in the Brillouin zone, varying sign etc..
A: Well if the photon interacts with something and something effectively slows it down. Then the photon will have mass. While the photon does not have any intrinsic mass any interaction will make it to have mass. This is similar to the case with gluons. Gluons can bundle up together. While the gluons themselves are massless, overall the glueball has mass due to the interaction between the color charged gluons.
