18. Prove by mathematical induction.

When $n=1$,
$$\begin{array}{l} y_{1}\left(z_{1}-z_{0}\right)=y_{1} z_{1}=0 \\ \Rightarrow\left\{\begin{array} { l } { y _ { 1 } = 0 , } \\ { z _ { 1 } = - x _ { 1 } } \end{array} \text { or } \left\{\begin{array}{l} y_{1}=x_{1}, \\ z_{1}=0 . \end{array}\right.\right. \end{array}$$

If $x_{1} \geqslant 0$, then $\left\{\begin{array}{l}y_{1}=x_{1} \\ z_{1}=0 ;\end{array}\right.$
If $x_{1}<0$, then $\left\{\begin{array}{l}y_{1}=0, \\ z_{1}=-x_{1} .\end{array}\right.$
Therefore, when $n=1$, the proposition holds.
Assume that when $n=k(k \geqslant 1)$, the proposition holds.
Then when $n=k+1$, the problem is equivalent to
$$\left\{\begin{array}{l} y_{k+1}-\left(z_{k+1}-z_{k}\right)=x_{k+1}+z_{k}, \\ y_{k+1}\left(z_{k+1}-z_{k}\right)=0, \\ y_{k+1} \geqslant 0, \\ z_{k+1}-z_{k} \geqslant 0, z_{0}=0 \end{array}\right.$$
$$\begin{array}{l} \Rightarrow\left\{\begin{array}{l} y_{k+1}=0, \\ z_{k+1}-z_{k}=-\left(x_{k+1}+z_{k}\right) \end{array}\right. \text { or } \\ \left\{\begin{array}{l} y_{k+1}=x_{k+1}+z_{k}, \\ z_{k+1}-z_{k}=0 . \end{array}\right. \end{array}$$

If $x_{k+1}+z_{k} \geqslant 0$, then $\left\{\begin{array}{l}y_{k+1}=x_{k+1}+z_{k}, \\ z_{k+1}=z_{k} ;\end{array}\right.$

Hence, when $n=k+1$, the proposition holds.
By mathematical induction, the proposition holds for any natural number $n$.