Elimination with 2 Bromide groups

Yes, this is an E2-type of elimination. The reactive conformation is that in which the two C-Br bonds are mutually trans, which sets them up for concerted elimination of Br₂ via an anti, coplanar transition state. Iodide ion functions as a nucleophile to attack one of the bromine atoms, which at the time of reaction is polarized toward Br(delta+) [positive halogen, electrophilic; the leaving group is polarized toward Br(delta -)]. The dehalogenation reaction proceeds through a transition state analogous to that associated with concerted anti, coplanar E2 elimination. The products are the corresponding alkene, IBr [polarization I(delta +) Br(delta -)]and bromide ion.

For additional information see Goering, H. L.; Espy, H. H. J. Amer. Chem. Soc. 1955, 77. 5023-5026.

Hope this helps.

You need to look carefully at the electron movements in the mechanism and less attention to the term "E2 elimination". We often use E2-elimination to refer to a dehydrohalogenation. This is a different reaction. In this case iodide is an electron donor. The product will also contain an I₃^- ion. You would get the similar reaction if you did a reductive elimination with a dissolving metal.

Thanks for the *delete me*  Now, the question asked for the major product.  Iodide is a good nuleophile and weak base, and from what I've read this should favour Sn2 over E2.  Why is the E2 product the major one?