First, whether a class is pseudoeffective or not depends only on its numerical equivalence class.  (The pseudoeffective cone is the closure of the cone of big classes, and $D$ is big if and only if $nD$ is numerically equivalent to $A + E$, where $A$ is an ample divisor on $X$, $E$ is an effective divisor on $X$ and $n > 0$ is an integer.)

Now suppose a class $\beta$ was such that both $\beta$ and $-\beta$ were pseudoeffective.  Then $\beta \cdot a_1 \cdots a_k \geq 0$ and $\leq 0$ for any ample classes $a_1, \dots, a_k$.  So $\beta \cdot a_1 \cdots a_k = 0$.  But any divisor class can be written as a difference of ample classes, so $\beta \cdot \delta_1 \cdots \delta_k = 0$ for all divisor classes $\delta_1, \dots, \delta_k$.  Thus, $\beta = 0$.

The standard reference for this material is Lazarsfeld's [Positivity in Algebraic Geometry][1].  I don't have a copy handy, but I'm $> 99\%$ certain that this fact is proved somewhere in Volume I.


  [1]: http://www.ams.org/journals/bull/2006-43-02/S0273-0979-06-01087-1/S0273-0979-06-01087-1.pdf