59th Ukrainian National Mathematical Olympiad

Let us rewrite the problem the following way: since $1001$ is divisible by $7$, then $$\begin{array}{rl}\overline{ab0cd}& =1000\cdot \overline{ab}+\overline{cd}=1001\cdot \overline{ab}+(\overline{cd}-\overline{ab})\\ & \Rightarrow (\overline{cd}-\overline{ab})\text{ is divisible by }7\\ & \Rightarrow 10(c-a)+(d-b)\text{ is divisible by }7\\ & \Rightarrow 3(c-a)+(d-b)\text{ is divisible by }7.\end{array}$$

Similarly, $3(d-b)+(a-c)$ must be divisible by $7$. Thus, $3x+y=7k$ and $3y+x=7l\Rightarrow 9x+3y=21k\Rightarrow 8x=7(3k-l)\Rightarrow x$ is divisible by $7$, and similarly $y$ is divisible by $7$. Therefore, there exist two pairs of nonzero digits, difference of which is divisible by $7$. There are two such pairs: $(8,1)$ and $(9,2)$. Clearly, the pair $(7,0)$ is not considered, since it contains zero, that already is one of the digits of the number. So, there are eight numbers that satisfy the conditions: $12089$, $19082$, $21098$, $28091$, $89012$, $82019$, $98021$ and $91028$. Their average equals $55055$.