20. Let $S_{k}=\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{k}}{a_{k+1}}$, it can be proven that for any $M>0$, there exists a positive integer $k_{0}$ such that when $k \geqslant k_{0}$, we have
$$S_{k}<k-M$$

From this, we can conclude that (1) and (2) both hold.
In fact, since $a_{1}, a_{2}, \cdots$ is strictly increasing, we have
$$k-S_{k}=\sum_{i=1}^{k} \frac{a_{i+1}-a_{i}}{a_{i+1}} \geqslant \frac{1}{a_{k+1}} \sum_{i=1}^{k}\left(a_{i+1}-a_{i}\right)=1-\frac{a_{1}}{a_{k+1}}$$

Since $a_{1}, a_{2}, \cdots$ is unbounded, we have $\lim _{k \rightarrow \infty} a_{k}=+\infty$, thus there exists $k_{1}$ such that when $k \geqslant k_{1}$, $\frac{a_{1}}{a_{k+1}}<\frac{1}{2}$.

Therefore, for any $k \geqslant k_{1}$, we have
$$S_{k}<k-\frac{1}{2}$$

Let $b_{1}=a_{k_{1}+1}, b_{2}=a_{k_{1}+2}, \cdots$, then the sequence $\left\{b_{k}\right\}$ is also a monotonic and unbounded sequence of positive numbers. From (2), we know there exists $k_{2}$ such that when $k \geqslant k_{2}$,
$$\frac{b_{1}}{b_{2}}+\frac{b_{2}}{b_{3}}+\cdots+\frac{b_{k}}{b_{k+1}}<k-\frac{1}{2}$$

Thus, for any $k \geqslant k_{1}+k_{2}$, we have $S_{k}<k-1$.
By continuing this process, (1) holds.