96 Let the sequence $\left\{a_{1}, a_{2}, \cdots,\right\}=\left\{\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \cdots\right\}$, then the 1988th term of this sequence $a_{1988}=$ $\qquad$ .
$96 \frac{39}{25}$. Group the sequence where the sum of the numerator and denominator is equal as a segment:
$$
\left(\frac{1}{1}\right),\left(\frac{2}{1}, \frac{1}{2}\right),\left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \ldots \ldots
$$

Let $a_{1}$ be the 1988th term located in the $n$th segment, then we have
$$
\frac{1}{2} n(n-1)<1988 \leqslant \frac{1}{2} n(n+1),
$$

Solving this, we get $n=63$. The first 62 segments contain
$$
\frac{1}{2} \times 62 \times 63=1953 \text{ (terms) },
$$

and in the 63rd segment, the sum of the numerator and denominator is 64, hence $a_{1988}=\frac{29}{35}$.