Proof for the forward direction: Let $p(x)$ denote a complex-coefficient polynomial satisfying the given conditions. If the multiple of $p(x)$ has no real positive roots, then $p(x)$ has no such roots either, so we need only show that the multiple has no positive real roots. This result follows directly from Descartes's Rule of Signs and the fact that all coefficients are positive. Thus, $p(x)$ has no positive real roots either.

Does the polynomial $p(x) = (x-i)(x-i)$ contradict the converse?

Pay attention, mathisfun! $Q(x)$ is a multiple of $P(x)$ if exists a polinomial $R(x)$ s.t. $Q(x)=P(x)R(x)$

Sorry! I did not know that the multiplier could be nonconstant. There goes my solution...

I have, I think, part of the proof for the reverse direction; someone out there could definitely improve or finish it: Assume WLOG that all roots are complex. Write $p(x)$ in factored form: $p(x) = (x-\alpha_1)(x-\alpha_2)...(x-\alpha_n)$ Let $\alpha_i'$ denote the complex conjugate of $\alpha_i$. $(x-\alpha_1')...(x-\alpha_n')p(x) = (x-\alpha_1)(x-\alpha_1')(x-\alpha_2)(x-\alpha_2')...(x-\alpha_n)(x-\alpha_n') \\ \\ = (x^2-(\alpha_1 + \alpha_1')x^ + |\alpha_1|^2)...(x^2-(\alpha_n + \alpha_n')x+|\alpha_n|^2)$ Clearly this polynomial has real coefficients. However if the real parts of some of the roots are positive, then the polynomial will have negative coefficients. I am still attempting to find a suitable factor to negate those negatives.

Just one of the old shortlist problems that solves this problem instantly : If $f(x)$ is a real polynomial such that $f(x)>0$ for all $x\geq0$ we have for some $n$ the coefficients of $(1+x)^nf(x)$ are positive.