Let the first number be the $n$ digit number $10^{n}\times n_1+10^{n-1}\times n_2+10^{n-2}\times n_3\cdots 10^{0}\times n_n$ and the second number be the $m$ digit number $10^{m}\times m_1+10^{m-1}\times m_2+10^{m-2}\times m_3+\cdots 10^{0}\times m_m$. The way for the first digits to be the same is $\iff \forall \max(m_1, n_1) \text{ has y digits}, n_1\pmod{y}\equiv m_1\pmod{y}$. However, we note that when the digits of $m_1$ minus the digits of $n_1$ is greater than $1$, or the digits of $n_1$ minus the digits of $m_1$ is greater than $1$, that either $m_1$ will be greater than $n_1$ or $n_1$ will be greater than $m_1$. Therefore they must have the same amount of digits. We now look at the unit digit of both numbers. For the unit digit of the first number to be the same as the unit digit of the second number, because they have the same amount of digits, we must have the unit digits are equal of both numbers. However, $20$ more units after that will be $\equiv 0\pmod{10}$ more unit digits, and $21$ more units after that will be $\equiv 1\pmod{10}$ unit digits after that. Therefore the unit digits cannot be the same in the last digit. The only case where this may be possible is if after $21$ more digits added, we go up to the next thousand, but in this case then the $21$ consecutive numbers or $20$ consecutive numbers will have a different amount of digits. We can either have that both unit digits $\pmod{10}$ are not equal, or they have a different amount of digits, therefore the numbers are $\boxed{\textbf{Not equal}}$.