Below is a summary of the discussion in Lazarsfeld's *Positivity in algebraic geometry I*, Ex. 1.4.7, Lem. 1.5.4, and Rmk. 1.5.6.

> **Lemma.** *Let $D$ be an $\mathbf R$-divisor on an abelian surface $A$. Then the following are equivalent:*
>
> 1. *$D$ is nef;*
> 2. *$D$ is pseudo-effective;*
> 3. *$D^2 \geq 0$ and $D \cdot H \geq 0$ for any ample divisor $H$.*

*Proof.* Implication 1 $\Rightarrow$ 3 is clear. For 2 $\Rightarrow$ 1, it suffices to treat the case where $D$ is effective and irreducible. Any translate $D + a$ for $a \in A$ is algebraically equivalent to $D$, so $D^2 = D(D + a) \geq 0$ as $D \neq D + a$ for $a \in A$ general. Finally, for 3 $\Rightarrow$ 2 it suffices to show that if $D$ is an integral divisor with $D^2 > 0$ and $D \cdot H > 0$, then some multiple of $D$ is linearly equivalent to an effective divisor. This follows from Riemann–Roch for abelian surfaces. (In fact $D$ is ample; see e.g. [this post](https://mathoverflow.net/a/141960/82179), or Prop. 1.5.17 in Lazarsfeld.) $\square$

**Example.** For example, if $E$ is an elliptic curve with CM in $\mathbf Z[\sqrt{-n}]$ for $n > 0$, and $\Delta, \Gamma \subseteq E \times E$ are the diagonal and the graph of "multiplication by $\sqrt{-n}$" respectively, then the matrix of the intersection form with respect to the basis $(h,v,\Delta,\Gamma)$ is
$$\begin{pmatrix}0 & 1 & 1 & n \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1+n \\ n & 1 & 1+n & 0 \end{pmatrix}.$$
A better basis is $(h+v,h-v,\Delta-h-v,\Gamma-h-nv)$, which gives the matrix
$$\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2n \end{pmatrix}.$$
With respect to this basis, the equations become
\begin{align*}
a^2 \geq b^2 + c^2 + nd^2, & & a \geq 0
\end{align*}
for a divisor $D = a(h+v) + b(h-v) + c(\Delta-h-v) + d(\Gamma-h-nv)$. These describe a circular cone in $\operatorname{NS}(A)_{\mathbf R} \cong \mathbf R^4$, so you can produce many effective classes close to the boundary with negative $\Delta$ or $\Gamma$ components. For example $(a,b,c,d) = (2m^2+1,2m^2,-2m,0)$ gives a divisor $D$ with
$$D^2 = 2\Big((2m^2+1)^2 - (2m^2)^2 - (-2m)^2\Big) = 2\Big( 4m^4 + 4m^2 + 1 - 4m^4 - 4m^2 \Big) = 2,$$
so $D$ is effective (even ample). Its coefficient in $\Delta$ is $-2m$.

**Remark.** I don't know if every pseudo-effective class is algebraically equivalent to an effective one. (This is certainly false for "linearly equivalent", as can be seen with $p \times E - q \times E$ for different points $p, q \in E$.) On a general abelian surface I don't expect this to be true, because [every effective class is ample if $A$ is simple](https://mathoverflow.net/q/25826/82179), but I imagine there might be classes on the boundary of the nef cone (if $A$ has complex multiplication).

For a product of isogenous CM elliptic curves, there is a little more hope.