For simplicity, consider symmetric matrices (but otherwise is fine too).
A real symmetric matrix $A$ is positive definite if for every real non-zero vector $x$: \begin{gather*} x^T A x > 0 \end{gather*} where $x^T A x$ is a scalar. Reverse the condition for negative definiteness.
Given this, there should be three conditions to satisfy: \begin{gather*} x^T A x > 0\\ x^T B x > 0\\ x^T (A+B)x < 0 \end{gather*}
I'm not sure how to proceed from here to construct an example of $A$ and $B$ which satisfy this.
No, it is not possible to find such matrices $A$ and $B$. Note that by linearity $$x^T (A+B)x=x^T Ax+x^T Bx.$$ Therefore if $x^T Ax>0$ and $x^T Bx\geq 0$ then $x^T (A+B)x>0$.