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I do not know the case where answer in general.

However, when $X$ is a surface and $Z$ is a curve, it is not difficult to see that the only possibility, in the answer case you are interested in, is easy:

$D=\mathcal{q}F$,

where $F$ is the class of a fiber of $f \colon X \to Z$ and $q \in \mathbb{Q}$.

In fact, since we are assuming that $-D$ is $f$-nef, we must have $-DF \geq 0$. Writing $D=D_1+D_2$, where $D_1$ is the maximal subdivisor contained in fibres of $f$, we obtain $D_1F=0$, $D_2F > 0$, hence

$D=D_1$ and $D_2= \emptyset$,

in particular $DF=0$ and $D^2 \leq 0$ by Zariski lemma.

Now assume that $-D+nF$ is nef for some $n \geq 1$; then

$0 \leq (-D+nF)^2=D^2$,

thus $D^2=0$. Again by Zariski lemma, $D$ must be a rational multiple of $F$.

1

In the case where $X$ is a surface and $Z$ a curve, the answer is easy:

$D=\mathcal{q}F$,

where $F$ is the class of a fiber of $f \colon X \to Z$ and $q \in \mathbb{Q}$.

In fact, since we are assuming that $-D$ is $f$-nef, we must have $-DF \geq 0$. Writing $D=D_1+D_2$, where $D_1$ is the maximal subdivisor contained in fibres of $f$, we obtain $D_1F=0$, $D_2F > 0$, hence

$D=D_1$ and $D_2= \emptyset$,

in particular $DF=0$ and $D^2 \leq 0$ by Zariski lemma.

Now assume that $-D+nF$ is nef for some $n \geq 1$; then

$0 \leq (-D+nF)^2=D^2$,

thus $D^2=0$. Again by Zariski lemma, $D$ must be a rational multiple of $F$.