I am re-editing this response because I got it wrong the first time around.
The following result, due to myself and José Barbosa Gomes, is (a small) part of the paper Periodic solutions of Hilbert's fourth problem.
Theorem. If a C2 reversible Finsler metric $F$ on a compact, connected Lie group which is neither $SU(2)$ nor $SO(3)$ has the same unparametrized geodesics as a bi-invariant Riemannian metric, then it is a bi-invariant metric.
So the answer to your question is yes for $N > 2$. Let me recall that a reversible Finsler metric is one for which the length of each tangent vector $v_x$ equals the length of its opposite $-v_x$.
If you admit the metric to be smooth (= as smooth as needed with the smoothness depending on the dimension) I can give you a sketchy shortcut to the proof in the paper.
First one proves that a smooth reversible Finsler metric whose geodesics are straight lines on a torus of dimension $k$ $(k > 1)$ is flat (= locally isometric to a normed space). This is done by looking a bit under the hood at the Busemann-Pogorelov solution of Hilbert's fourth problem.
Now if $G$ is neither $SU(2)$ nor $SO(3)$, then it has rank greater than one as a symmetric space when provided with a bi-invariant Riemannian metric. This means every geodesic belongs to a flat torus of dimension at least two. If you have a Finsler metric $F$ with the same unparametrized geodesics, then those tori will be totally geodesic for $F$ and, by the result above, the restriction of F to those tori will also be a flat metric. From this we deduce that $F$ is not just projectively equivalent to the bi-invariant Riemannian metric, but also affinely equivalent to it (i.e., the midpoints of geodesic segments are the same). By work of Z.I. Szabo (http://front.math.ucdavis.edu/0601.5522), this is known to imply that $(G,F)$ is a symmetric space and so that the metric $F$ is bi-invariant.