Nice exercise, though I doubt that it can be considered as a research question. Put $y=x^2$, and write
$q(x)= a(y)+xb(y)$. I claim that the polynomials $a$ and $b$ are coprime: if a non-constant polynomial $f$ divides $a$ and $b$, then $f(x^2)$ divides $q(x)$, but that implies that some of the roots $x_i$ of $q$ are negative. Note also that since the roots are positive $y$ does not divide $a$. Therefore
there exist polynomials $c,d$ such that $a(y)c(y)+yb(y)d(y)=\dfrac{1}{2} $. Put $p(x)=c(x^2)+xd(x^2)$; then $q(x)p(x)=\dfrac{1}{2}+ x g(x^2) $, with $g= ad+bc$, so $q$ answers the question.

As usual you can replace $c(y),d(y)$ by $c(y)-yb(y)e(y), d(y)+ a(y)e(y)$ for some polynomial $e$, so the solution is far from unique.