One is more stable according to its charge. The other one does not satisfy as much as the first does the formal charge rule but has a complete octet.
2 Answers
Beryllium chloride has a significant ionic contribution to its electronic structure. A quick calculation, DF-BP86/def2-SVP, reveals that the charge of beryllium is $\mathbf{q}(\ce{Be})=0.8$ and of chlorine $\mathbf{q}(\ce{Cl})=-0.4$, based on natural bond orbital analysis.
The Lewis structures best describing this are depicted below. The percentage indicates how much Lewis character this model actually describes based on the before mentioned calculation.
Of course this is just an approximation and the truth lies somewhere in between, like Eljee already mentioned. For these bonding situation, resonance is unavoidable. In quantum chemistry these phenomena are usually very well described by valence bond theory.
Another possibility, which is quite complementary, is looking at the structure with molecular orbital theory. Since the molecule has $D_\mathrm{\infty{h}}$ symmetry, you can observe $\pi$ bonds, that are delocalised over the whole molecule. And these are only depicted in the right hand side structure above. Below are the valence orbitals of $\ce{BeCl2}$, DF-BP86/def2-SVP.
So you can see, that one Lewis structure is not enough to actually describe the bonding in that molecule well, but as a first approximation, the left hand structure above should be sufficient.
The two structures drawn are both contributing to a total distribution of the charge. These are called Resonance forms. In some cases it means the electrons are equally distributed over a molecule, due to symmetry. In the case of your molecule the contribution of the two forms will most likely not be equal, but the truth lies somewhere in between the two structures you have drawn.