@coco007wind I have a proof for n >= 3.

We can just consider the following two numbers: 6*10^{n-1}+6 and 4*10^{n-1}+7, which are both n digits.

We have (6*10^{n-1}+6)*(4*10^{n-1}+7) = 24*10^{2*n-2} + 66*10^{n-1} + 42, which is 2n digits.

Also, when n >= 3, i.e., n-1>=2, we have 24*10^{2*n-2} + 66*10^{n-1} + 42 like 240..0660..042, which is palindrome, here the number of 0s between is just n-3.