Answer
No definite parity - neither even nor odd.
Work Step by Step
We have that $E(-x)=E(x)$ and $O(-x)=-O(x)$.
Let $\Phi=E+O$. Then
$$\Phi(-x)=E(-x)+O(-x)=E(x)-O(x)$$
which is different than both $\Phi(x)$ and $-\Phi(x)$ so this function is neither even nor odd.