Thomas Bauer shows in http://arxiv.org/pdf/alggeom/9712019v1.pdf that for a complex abelian variety a nef line bundle is numerically equivalent to an effective divisor (this is shown in Lemma 1.1). It seems to me (by comments in other papers) that this is known over a general algebraically closed field, but I have yet to find a proof. Since Bauer's proof deals with the eigenvalues of the Hermitian form associated to the (polarized) abelian variety, I'm not quite sure how a proof would proceed over an arbitrary algebraically closed field. Any suggestions or references?
Here is a sketch of a purely algebraic proof based on the theory developed in Chapter 3 of Mumford's "Abelian Varieties". Let $L$ be a nef line bundle on the abelian variety $A$ of dimension $g$. If $K(L)$ is finite, then the index of $L$ is a well defined integer between $0$ and $g$ and we need to show that $g=0$. By the theorem on p. 155, the index is the number of positive roots of the polynomial $P(n) = \chi(L \otimes M^n)$, where $M$ is any ample line bundle. Using RiemannRoch and the nefness of $L$ one sees that all the coefficients of the polynomial are nonnegative so it has no positive roots. We now reduce the general case to the one above. We may assume, by replacing $L$ by $L \otimes (1)^*L$, that $L$ is symmetric. If $K(L)$ is not finite, i.e. $T_x^*L \cong L$ for $x$ in a positive dimensional subvariety $B$ of $A$ we claim that $L^{2}$ descends to a line bundle $L'$ on $A/B$. This follows as we can get descent data for $L^2$ by choosing symmetric isomorphisms $T_x^*L \cong L$ for all $x \in B$. (Note that one cannot always descend $L$ as shown by the example of a line bundle of order $2$ in $Pic(A)$.) Since $L$ is nef it follows that $L'$ is also nef, so it is numerically equivalent to an effective line bundle by induction. (In the sketch above, the class of $L$ modulo numerical equivalence is replaced with a multiple. One can perhaps avoid this by an additional argument.) 

