Example of nef divisor with Iitaka dimension 0

Here's an example that works over an uncountable field.

Let $E$ be an elliptic curve in $\mathbf P^2$ and let $p_1, \ldots, p_9$ be 9 points on $E$. Blowing up these points, the proper transform $C$ of $E$ is an irreducible curve with self-intersection 0, hence nef.

Notice that if some multiple $mC$ of $C$ moves, it must move to a curve disjoint from $C$. But that implies that the corresponding power $N_C^m$ of the normal bundle of $C$ is trivial.

Now by choosing the points $p_1, \ldots, p_9$ on $E$ appropriately, we can arrange that $N_C$ is any desired element of $\operatorname{Pic}^0(C)$. In particular since our field is uncountable we can choose it to be a non-torsion element, so no power $N_C^m$ is trivial, and therefore no multiple $mC$ moves.

This is not exactly what was asked, but I think it is close to the spirit of the question: Mumford's example of a non-ample divisor that is positive on every curve gives an example of a numerically non-trivial nef divisor with Iitaka dimension $-1$.

Here is a sketch, you can find a complete proof in many places, or fill in the gaps yourself:

Let $B$ be a curve of genus at least $2$. Then there exists $\mathscr E$, a locally free sheaf on $B$ of rank $2$ and degree $0$ such that $\operatorname{Sym}^m(\mathscr E)$ is stable for all $m>0$. (Proof: HW).

Let $X=\mathbb P(\mathscr E)$ and $D$ the divisor corresponding to $\mathscr O_{\mathbb P(\mathscr E)}(1)$. Then $D\cdot C>0$ for any effective curve $C\subseteq X$. (Proof: HW. Hint: use the stability assumption).

So, $D$ is nef, in fact as nef as it can be $\overset{..}\smile$.

However, $D$ is not ample, because $D^2=0$. (Proof: $\deg\mathscr E=0$).

This is the usual reason this example is mentioned, that is, that being positive on curves is not enough for being ample. But it actually gives an example of a numerically non-trivial nef divisor that does not have any effective representative, because if $D$ was represented by an effective divisor, then it would be positive on it, which would contradict that $D^2=0$. (And, in fact in that case, $D$ would be ample. In other words, any non-ample divisor on a surface that is positive on every curve must have an empty linear system).