Palindrome of numbers from 1-n

No. With $n=1$, we do have a palindrome of course. But for $n>1$ we can clearly exclude the case $n\le 10$. In fact, we need $n\equiv 1\pmod {10}$, as the palindromic string ms must end in "$\ldots 1$". Let $k\ge1$ with $10^k<n<10^{k+1}$. Then there is exactly one position in the assumed palindromic string "$12345\ldots54321$" where "$1\underbrace{0\ldots0}_k1$" occurs: In the absence of leading zeroes, only the two $1$s in this block can be leading digits of some numbers, and since no number has more than $k+1$ digits, indeed both $1$s must be leading digits, i.e. the only position where this pattern occurs is at the number $10^k$ (together with the leading $1$ of $10^k+1$). For a palindrome, such a uniquely occuring subpalindrome must be in the very middle. But it is preceeded by "$\underbrace{9\ldots9}_k$" from $10^k-1$ and followed by "$\underbrace{0\ldots0}_{k-1}1$" as the rest of $10^k+1$, contradicting the palindromic symmetry.