I will say zero is effective on a complete, smooth variety $X$ if some positive linear combination of irreducible varieties is rationally equivalent to zero. In other words, zero is effective if there exist (not necessarily distinct) irreducible closed subvarieties $V_1, \dotsc, V_k$ of $X$, with $k \geq 1$, such that $$[V_1] + \dotsb + [V_k] = 0$$ in the Chow ring. (If you know how to answer this question with rational equivalence replaced by numerical or algebraic equivalence, please do so.)
If $X$ is projective, then zero is not effective on $X$ since any subvariety $V$ has positive degree after $X$ is embedded in $\mathbb P^n$.
[More intrinsically, no positive linear combination of points can be rationally equivalent to zero; and if $D$ is an ample divisor on $X$ and $V$ is a dimension-$k$ variety in $X$, then $[D]^k . [V]$ is a positive linear combination of points, hence nonzero in the Chow ring; whereas $[D]^k .  = 0$ in the Chow ring.]
At one point, I assumed naively that zero would not be effective on any complete smooth variety; the requirement that $X$ be projective was surely an artifact of the proof. However, there is the example of a non-projective complete smooth threefold (due to Hironaka) that is described in Hartshorne Appendix B Example 3.4.1 is shown to be non-projective precisely by producing sum of two curves on the variety rationally (or at least algebraically) equivalent to zero--showing that on this variety, zero is (algebraically) effective. [Note that Hironaka's construction can easily be performed so as to ensure rational effectiveness.]
The next most naive conjecture would be that the effectiveness of zero characterizes non-projective smooth complete varieties among all smooth complete varieties. If true, this would give what I consider a nicer characterization for projectivity than simply the existence of an ample divisor. Philosophically, it seems that zero should not be effective on any "nice" variety, so a characterization like this would make it somehow less annoying when proofs use a hypothesis of projectivity and not only completeness.
Chances are the statement "next most naive" conjecture in the preceeding paragraph is false. However, even if I could somehow invent a possible counterexample, I would have no idea how to prove it was not projective, since the only way I know to give such a proof is to show that zero is effective.
Hence, the question:
Does there exist a smooth, complete, non-projective variety on which zero is not effective (in the sense of being numerically equivalent to an effective algebraic cycle)?