Solution:

If we cross multiply, we obtain $a_{n} a_{n-1}=n a_{n-1}+(n-1) a_{n}$, which we can rearrange and factor as $\left(a_{n}-n\right)\left(a_{n-1}-(n-1)\right)=n(n-1)$.

Let $b_{n}=a_{n}-n$. Then, $b_{n} b_{n-1}=n(n-1)$. If we let $b_{1}=t$, then we have by induction that $b_{n}=n t$ if $n$ is odd and $b_{n}=n / t$ if $n$ is even. So we have
$$a_{n}= \begin{cases}n t+n & \text{ if } n \text{ odd } \\ n / t+n & \text{ if } n \text{ even }\end{cases}$$
for some real number $t$. We have $20 / t+20=23 t+23$, so $t \in\{-1,20 / 23\}$. But if $t=-1$, then $a_{1}=0$ which is not positive, so $t=20 / 23$ and $a_{100}=100 / t+100=215$.