I stumbled across this paper (http://www.willamette.edu/~knyman/papers/Fan_Sperner.pdf) and am now trying to understand "Fan's N+1 Lemma" which is a generalization of Tucker's Lemma. However, even for the case $n=1$ I fail to find a triangulation of $\Sigma^n$ that satisfies the condition of the lemma.
Did I understand correctly that a symmetric triangulation has anti-symmetric labelling if $l(-v)=-l(v)$ for not only the vertices of $\Sigma^n$ but also all the vertices that result from the triangulation?
For the case $n=1$ the surface $\Sigma_1$ is just the boundary of a square (rotated by 45 degrees but that's immaterial). Here the triangulations are $1$-simplices, in other words edges. The simplest "triangulation" respecting the orthants is just the four vertices and edges of the square itself. It's easy to give this an antisymmetric $2$-labelling without complementary edges: just give one pair of opposite vertices $\pm1$ and the other pair $\pm2$.
A finer triangulation in this case would just be further subdividing the edges by introducing new intermediate vertices, but not cutting through the interior.
You didn't give any details in the question I suspect you had the wrong initial picture of the triangulation, perhaps thinking that the interior of the square is divided into four triangles around a central vertex. In such a case, yes of course no actual triangle could avoid a complementary edge using distinct labels from $\{\pm 1, \pm 2\}$.