Sorry for the delay, I was waiting for a response from a string theorist I was talking to:
"Sorry for the delayed reply, I have been trying to think of the best way to explain it. What you end up getting is what is known as a tower of excitations, simply put
m^2 = (\frac{n\pi}{L})^2
Where L is a length related to compactification, ie its a small number and n=0,+/- 1,+/-2... so that its the n=0 mode that gives you a massless particle in string theory. This isn't anything fundamental though.
For the next bit you will need tex the world add on for your browser.
In general we require gauge bosons to be massless, regardless if we are doing String Theory, GR, or any other field theory. Consider E&M, its Lagrangian density is
\mathcal{L}=F ^{\mu\nu} F_{\mu\nu}
Where F is the field strength tensor
F_{\mu\nu}= \partial_\mu A_\nu-\partial_\nu A_\mu
After some algebra you get
F_{\mu\nu} F^{\nu\mu}=\partial_\mu A_\nu \partial ^\mu A ^\nu -\partial_\nu A_\mu \partial ^\mu A ^\nu
Schematically this is just
F ^2 = (\nabla A) ^2
Which looks like part of the lagrangian that leads to the Klein Gordon equation. The crucial part that it is missing is the m2 A2 term. What happens if we put that in by hand? Its not good. A is the Gauge field, its the photon when we quantize the theory. Since its the gauge field it must obey gauge symmetry, for abelian fields this means that the lagrangian must be invariant under the transformation
A_\mu\rightarrow A_\mu -\partial_\mu f
where f is an arbitrary function. When we add a mass term to our original lagrangian it loses gauge invariance, therefore all gauge fields, by definition, have to be massless. The higgs mechanism allows a way around this for low energies, but that is another discussion entirely."