Is there a positive integer $N>10^{20}$ such that all its decimal digits are odd,
the numbers of digits 1, 3, 5, 7, 9 in its decimal representation are equal,
and it is divisible by each 20-digit number obtained from it by deleting
digits? (Neither deleted nor remaining digits must be consecutive.)
The answer is affirmative.
We'll construct $N>10^{20}$ so that $A\mid N$ for all $20$-digit number $A$ consists of only digits $1,3,5,7,9$.
Among those $A$s, let $A_0$ be (one of) number with highest $\nu_5 (A)$.
Then let $A_0'$ be the number formed by placing some digits $1,3,5,7,9$ on the left of $A_0$ so that there're equal no. of each digit.
It's clear that $5^{\nu_5 (A)}\mid A_0'$ for all $20$-digit number $A$ consists of only digits $1,3,5,7,9$.
Then, for each $20$-digit number $A$ consists of only digits $1,3,5,7,9$, there exists positive integer $t_A$ that $$\frac{A}{5^{\nu_5 (A)}} \mid \overline{\underbrace{A_0'A_0'...A_0'}_{t_A}},$$which exists by Pigeonhole and note that $\gcd ( \frac{A}{5^{\nu_5 (A)}},10) =1$.
Finally, choose $N=\overline{\underbrace{A_0'A_0'...A_0'}_{T}}$ where $T=\mathrm{lcm}{(t_A)}$.