As $207 = 9 \cdot 23$ and $9$ and $23$ are relatively prime, it suffices to find a difference that would be divisible by both $9$ and $23$.

All the $20$-digit numbers listed are divisible by $9$ because the sum of their digits is $90$. Therefore the difference of any two of them is also divisible by $9$.

It remains to show that the difference of some two of them is divisible by $23$. For this note that there are $24$ different orderings of $3$, $4$, $5$ and $6$. Therefore there are $24$ numbers written in total, but there are $23$ different possible remainders. So there must exist some two among those that give the same remainder when divided by $23$. Their difference is divisible by $23$.