Ukrainian National Mathematical Olympiad

Let an arbitrary natural number $n$ and consider all palindromic numbers that have exactly $n$ digits. We consider $n$ odd and $n$ even separately.

When $n = 2m$, palindrome becomes $a_1a_2\dots a_{m-1}a_m a_m a_{m-1}\dots a_2a_1$, the only restriction on its digits is $a_1 \neq 0$. Hence there are $9 \cdot 10^{m-1}$ of them and each is greater than $10^{2m-1}$. Therefore the sum of reciprocals is not greater than $\frac{9 \cdot 10^{m-1}}{10^{2m-1}} = \frac{9}{10^m}$. For all $m$ the sum is not greater than
$$\sum_{m=1}^{\infty} \frac{9}{10^m} = 9 \cdot \sum_{m=1}^{\infty} \frac{1}{10^m} = 9 \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}} = 9 \cdot \frac{1}{9} = 1.$$

When $n = 2m-1$, palindrome becomes $a_1a_2\dots a_{m-1}a_m a_{m-1}\dots a_2a_1$, and again the only restriction on its digits is $a_1 \neq 0$. Hence there are $9 \cdot 10^{m-1}$ of them and each is greater than $10^{2m-2}$. Therefore the sum of reciprocals is not greater than $\frac{9 \cdot 10^{m-1}}{10^{2m-2}} = \frac{9}{10^{m-1}}$. For all $m$ the sum is not greater than
$$\sum_{m=1}^{\infty} \frac{9}{10^{m-1}} = 90 \cdot \sum_{m=1}^{\infty} \frac{1}{10^m} = 90 \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}} = 90 \cdot \frac{1}{9} = 10.$$

Therefore, the sum does not exceed $11$.