Introduce binary variable $x_{i,j}$ to indicate whether $\beta_{i,j}>0$ and linear constraints:
\begin{align}
\beta_{i,j} &\le x_{i,j}\\
x_{i,j} + x_{j,i} &\le 1
\end{align}
(The big-M here is 1.)
The first constraint enforces
$$\beta_{i,j}>0 \implies x_{i,j} = 1.$$
The second constraint enforces
$$x_{i,j} = 1 \implies x_{j,i} = 0.$$
The first constraint enforces
$$x_{j,i} = 0 \implies \beta_{j,i} \le 0.$$
The lower bound on $\beta$ enforces
$$\beta_{j,i} \le 0 \implies \beta_{j,i} = 0.$$
So
$$\beta_{i,j}>0 \implies \beta_{j,i} = 0,$$
as desired.