First, we prove a lemma:

Lemma 1. Suppose there is a $k$-digit positive integer $X$, $k > 2$. Then such a palindrome can be subtracted from it, that the remaining number consists of no more than $\lfloor \frac{1}{2}k+1 \rfloor$ digits.

Proof. Suppose the first $\lfloor \frac{k+1}{2} \rfloor$ digits of $X$ make up the number $A \ne 100\dots0$. Then $B$, a number formed from $(A-1)$ by writing the digits in the opposite order (i.e. $B$ can begin with several zeros). Then, for an even $k$, we subtract number $\overline{(A-1)B}$, and for an odd $k$, number $\overline{(A-1)B'}$, where $B'$ is number $B$ but without the first digit. Then after such subtraction, the first digit on the left, which may be non-zero, is $\lfloor \frac{1}{2}k+1 \rfloor$-th digit. In case if $A=100\dots0$, we simply subtract from $X$ number $99\dots9$, which has one fewer digits.

Lemma is proved.

Lemma 2. A number with no more than three digits can be represented as a sum of no more than three palindromes.

Proof. For two-digit and single-digit numbers, this is obvious. For the three-digit number $abc$, we consider the following cases.

For $a \le c$, $abc = aba + (c-a)$.

For $a = c+1$, $abc = cbc + 99 + 1$.

For $a > c + 1$, $abc = cbc + (a-c)00 = cbc + (a-c-1)9(a-c-1) + (11-a+c)$.

Lemma is proved.

Now let's calculate how the number of digits will change if we first take a $2020$-digit number, then apply lemma 1 ten times and then lemma 2:
$$2020 \rightarrow 1011 \rightarrow 506 \rightarrow 254 \rightarrow 128 \rightarrow 65 \rightarrow 33 \rightarrow 17 \rightarrow 9 \rightarrow 5 \rightarrow 3.$$
Thus, we get no more than $13$ palindromes. If the initial number has fewer digits, then the number of palindromes can only decrease.