[Solution] Let the value of the above complex fraction with $k$ fraction lines be $\frac{m_{k}}{n_{k}}\left(m_{k} 、 n_{k}\right.$ being coprime natural numbers), then
$$\frac{m_{k+1}}{n_{k+1}}=\frac{1}{1+\frac{m_{k}}{n_{k}}}=\frac{n_{k}}{m_{k}+n_{k}}$$

Notice that
$$\begin{array}{l} \frac{m_{1}}{n_{1}}=\frac{1}{1}, m_{1}=1, n_{1}=1 \\ \frac{m_{2}}{n_{2}}=\frac{n_{1}}{m_{1}+n_{1}}=\frac{1}{2}, \quad m_{2}=1, n_{2}=2 \\ \frac{m_{3}}{n_{3}}=\frac{n_{2}}{m_{2}+n_{2}}=\frac{2}{1+2}=\frac{2}{3} \\ m_{3}=2, n_{3}=3 \\ \ldots \ldots \ldots \ldots \end{array}$$

Therefore, $\frac{m_{k}}{n_{k}}=\frac{n_{k-1}}{m_{k-1}+n_{k-1}}=\frac{F_{k}}{F_{k+1}}$,
where $F_{k}$ is the $k$-th term of the Fibonacci sequence, i.e., $F_{1}=1, F_{2}=1, F_{k+2}=F_{k+1}+$
$$F_{k} \quad(k=1,2, \cdots)$$

Thus,
$$\begin{aligned} & m^{2}+m n-n^{2} \\ = & F_{1988}^{2}+F_{1988} F_{1989}-F_{1989}^{2} \\ = & F_{1988}^{2}+F_{1988}\left(F_{1988}+F_{1987}\right)-F_{1987}^{2} \\ & -2 F_{1988} F_{1987}-F_{1988}^{2} \\ = & -\left[F_{1987}^{2}+F_{1987} F_{1988}-F_{1988}^{2}\right] \\ = & F_{1986}^{2}+F_{1996} F_{1987}-F_{1987}^{2} \\ = & \cdots \cdots \cdots \cdots . \\ = & F_{2}^{2}+F_{2} F_{3}-F_{3}^{2} \\ = & 1^{2}+1 \times 2-2^{2} \\ = & -1 . \end{aligned}$$