Solution:
We look at the cycles formed by $\pi$. Let $\operatorname{ord}_{\pi}(n)$ denote the smallest $m$ such that $\pi^{m}(n)=n$. In particular, the condition implies that $\operatorname{ord}_{\pi}(20) \mid 20$ and $\operatorname{ord}_{\pi}(21) \mid 21$.

Claim 1. $20$ and $21$ cannot be in the same cycle.
Proof. If $20$ and $21$ were in the same cycle, then $x=\operatorname{ord}_{\pi}(20)=\operatorname{ord}_{\pi}(21)$ for some $x$. Then $x>1$ since the cycle contains both $20$ and $21$, but $x|20, x|21$ implies $x=1$, a contradiction.

Claim 2. The probability that $a=\operatorname{ord}_{\pi}(20), b=\operatorname{ord}_{\pi}(21)$ for some fixed $a, b$ such that $a+b \leq 100$ is $\frac{1}{99 \cdot 100}$.
Proof. We can just count these permutations. We first choose $a-1$ elements of $[100] \setminus \{20,21\}$ to be in the cycle of $20$, then we similarly choose $b-1$ to be in the cycle of $21$. We then have $(a-1)!$ ways to reorder within the cycle of $20$, $(b-1)!$ ways to reorder within the cycle of $21$, and $(100-a-b)!$ ways to permute the remaining elements. The total number of ways is just
$$\frac{98!}{(a-1)!(b-1)!(100-a-b)!} \cdot (a-1)!(b-1)!(100-a-b)! = 98!,$$
so the probability this happens is just $\frac{98!}{100!}=\frac{1}{9900}$.

Now, since $\operatorname{ord}_{\pi}(20) \mid 20$ and $\operatorname{ord}_{\pi}(21) \mid 21$, we have $6$ possible values for $\operatorname{ord}_{\pi}(20)$ and $4$ for $\operatorname{ord}_{\pi}(21)$, so in total we have a $\frac{6 \cdot 4}{9900}=\frac{2}{825}$ probability that the condition is satisfied.