Hope this helps, it doesn't give a definite answer, but it tells about where to find an obstruction.
First off, the existence of multiple fibers in an elliptic fibration is an obstruction to the existence of a differentiable section (over $\mathbb{C}$). On the other hand we can always get rid of the multiple fibers by passing to an \'etale cover.

Now a more elaborate answer, given an elliptic fibration with no sections $X\rightarrow B$ we can associate a fibration $J\rightarrow B$ which has a section and a rational map $\phi:J\times_BX\rightarrow B$ that commutes with projections to $B$ and has certain properties. The family $J$ is called *jacobian family*.

The elliptic fibrations are classified by their jacobian fibrations and here comes the change of quantifiers. One can introduce a group structure on the set $I(J)$ of elliptic fibrations with a given jacobian fibration. Hence if the class of the fibration $X\rightarrow B$ in $I(J)$ is not zero, (the obstruction) the fibration $X$ has no differentiable sections. This works the same way the the first Chern class of a line bundle $L$ gives and obstruction for $L$ to be trivial.

rationalelliptic surface has 240integralsections, forming an $E_8$ root system. Shioda found some beautiful geometry and formulas related to this picture and its degenerations, and there's probably still more to be said about that. Elliptic K3's are much more complicated; e.g. in characteristic zero the Néron-Severi group can't fill out $H^2$, because $h^{2,0}$ and $h^{0,2}$ are nontrivial, and generically there needn't be a section at all. – Noam D. Elkies Apr 5 '12 at 3:49