Triange-free graph and its complement has Lovász number > 3 I found an example by the method in the paper Explicit Ramsey graphs and orthonormal labelings by Noga Alon 1994. The graph is around $10^6$ vertices, anyone knows smaller graph which is Triangle-free and its complement has Lovasz number >3?
Lovász theta number: https://en.wikipedia.org/wiki/Lov%C3%A1sz_number
paper: http://www.emis.ams.org/journals/EJC/Volume_1/PDF/v1i1r12.pdf
 A: Define $H_{n,k}$ to be the Cayley graph for $\mathbb{Z}_2^n$ whose connection set consists of all elements containing $k$ 1's. If $k$ is odd then this graph is bipartite, and if $k$ is even and $k < n$, then it is not bipartite and has two isomorphic components. Also, this graph is easily seen to be arc transitive, and so Lovasz theta of its complement is equal to $1 - \frac{d}{\tau}$ where $d$ is its degree and $\tau$ is its least eigenvalue. Since it is an abelian Cayley graph, the eigenvalues are relatively easy to compute. This is done for $k = 1,2,3$ and certain $n$ in this paper. More generally, for $k + 1 \le n \le 2k-2$ and $k$ even, Lovasz theta of the complement of $H_{n,k}$ is equal to $\frac{2}{2-(n/k)}$. This is greater than 3 whenever $\frac{n}{k} > \frac{4}{3}$. It is easy to see that for even $k$ the graph $H_{n,k}$ is triangle-free if and only if $\frac{n}{k} < \frac{3}{2}$. Therefore, if $k$ is even and $\frac{4}{3} < \frac{n}{k} < \frac{3}{2}$, then $H_{n,k}$ is triangle-free and Lovasz theta of its complement is greater than 3. The smallest such example is when $n = 11$ and $k = 8$. This has 1024 vertices if you consider just one component.
Another example can be found among the Haggkvist-Hell graphs, studied here. The vertices of the graph $HH_r(n)$ are the ordered pairs $(h,T)$ where $h \in [n]$, $T \subseteq [n]$, $|T| = r$, and $h \notin T$. Two vertices $(h,T)$ and $(h',T')$ are adjacent if $h \in T'$, $h' \in T$, and $T \cap T' = \varnothing$. All of these graphs are triangle-free. Moreover, for
$$n \ge r\frac{(r-1)^3 - 1}{r-2} + 1,$$
any graph $G$ with maximum degree $r$ has a homomorphism to $HH_r(n)$ if and only if $G$ is triangle-free. If there is a homomorphism from some graph $G$ to some graph $H$, then Lovasz theta of the complement of $G$ is at most that of the complement of $H$. So for fixed $r$ and large enough $n$, the graph $HH_r(n)$ must have Lovasz theta of its complement greater than or equal to Lovasz theta of the complement of any triangle-free graph with maximum degree $r$.
