Brazil

Note that $N=\stackrel{m\text{ ones}}{\overbrace{111\dots 1}}\times \stackrel{n\text{ ones}}{\overbrace{111\dots 1}}$ has exactly $m+n-1$ digits, since $\stackrel{m\text{ ones}}{\overbrace{111\dots 1}}\times \stackrel{n\text{ ones}}{\overbrace{111\dots 1}}<2\cdot 10^{m-1}\times 2\cdot 10^{n-1}=4\cdot 10^{m+n-2}$. If $m,n>9$, then considering the tenth leftmost digit, there would be a carry, thus the number consisting of the first nine digits of $N$ is bigger than the number formed by the last nine digits of $N$, in reversed order. Then if $m$ and $n$ are bigger than $9$ the number $N$ is not a palindrome.

If one of the numbers $m$, $n$ does not exceed $9$ then there won't be a carry and thus the number $N$ is palindrome.