Problem 1737

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

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 $55055$.