As G.Myerson already noted we must assume $W$ is finite.
Then a different approach for the "warmup" is to observe that
if $\omega \in W$ is not a positive real then some power
$\omega^N$ ($N \geq 1$) has non-positive real part.  Therefore $\omega$
is a root of the polynomial $(X^N-\omega^N)(X^N-\overline\omega^N)$
which has only nonnegative coefficients.  The product over all $\omega\in W$
gives a nonnegative polynomial of the desired kind.  It also gives
an upper bound $3^{\left|W\right|}$ on the number of monomials.

If $\{{\rm Im}\log\omega \colon \omega \in W, \omega \neq 0\} \cup \{2\pi\}$
is ${\bf Q}$-linearly independent then a single $N$ will do for all
$\omega\in W$, because as $N$ varies the $\left|W\right|$-tuples of angles formed by
the $N$-th powers are dense in $({\bf R}/2\pi{\bf Z})^{\left|W\right|}$.
In that case a polynomial of degree $2\left|W\right|$ in $X^N$ will do,
and we get an upper bound $2\left|W\right|+1$.

For a lower bound, clearly $2$ is enough **iff** there is some $N$ such that
all nonzero elements of $W$ have the same $N$-th power and that power
is negative; else we need at least $3$ monomials.
If $W$ contains *all* $n-1$ nontrivial $n$-th roots of unity
then we need at least $n$ monomials, because a polynomial $\sum_k a(k) x^k$
is a multiple of $(x^n-1)\big/(x-1)$ **iff** the $n$ sums
$\sum_j a(k_0+nj)$ ($k_0=0,1,2,\ldots,n-1$) are all equal,
and at least one of them must be positive if the $a(k)$ are nonnegative
and not all zero.