Let's determine those five-digit numbers, which, when divided by the four-digit number obtained by omitting the middle digit, result in an integer.

 

Find the five-digit numbers such that when divided by the four-digit number obtained by omitting the middle digit, the result is an integer.
Let the five-digit number be denoted by $A$, and the four-digit number obtained by omitting the middle digit of $A$ be denoted by $B$. We are looking for a number such that $A$ is a multiple of $B$. Since the difference between multiples of $B$ is also divisible by $B$, $A-10 \cdot B$ is also a multiple of $B$. Since $A$ and $10 \cdot B$ start with the same two digits, their difference can be at most a three-digit number. However, this can only be a multiple of the four-digit number $B$ if its value is 0. Therefore, if $A$ is divisible by $B$, then $A=10 \cdot B$.

We know that the last two digits of $A$ and $B$ are the same. Thus, because $A=10 \cdot B$, $B$ must end in 0. Therefore, the tenfold of $B$, which is $A$, is also divisible by 100, but then this is also true for $B$, and thus $A$ is also divisible by 1000.

The obtained condition is already sufficient, since a five-digit number divisible by 1000 means that omitting the middle digit of $A$ is equivalent to dividing by 10. Therefore, the sought numbers are the five-digit multiples of 1000, that is, the five-digit numbers whose last three digits are 0.