I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no, based on this:

**Claim:** A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic. 

Since ergodic linear toral automorphisms have positive entropy (see for instance [here][1]), the claim answers your question negatively.

Proof of the claim:
The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair.
Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. This implies that the dynamics fibers over a circle: you can find an integer eigenvector with eigenvalue $1$, and after a linear change of coordinates you may assume the dynamics is $(x,y) \mapsto (x, T'y)$ for $(x,y)\in \mathbb{T}\times \mathbb{T}^{d-1}$. Clearly if you have a Li-Yorke pair, both points must belong to the same fiber, and the automorphism $T'$ of $\mathbb{T}^{d-1}$ also has a Li-Yorke pair. So you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.


  [1]: http://www.springer.com/us/book/9783540077978