7. At the beginning, there are 10 consecutive positive integers written on the blackboard. The following operation is performed on the numbers on the blackboard: take any two numbers $a$ and $b$, and replace them with the numbers $a^{2}-2011 b^{2}$ and $a b$. After several such operations, all 10 numbers originally on the blackboard have been replaced. Question: Is it possible that the 10 numbers on the blackboard are still 10 consecutive positive integers at this time?
7. Impossible.

Assume after several operations, we get 10 consecutive positive integers, and all the original 10 numbers have been replaced.
First, we give a lemma.
Lemma: For any given positive integer $k$, the number of multiples of $k$ among the 10 numbers will not decrease after the operation.
This lemma is obvious.
Returning to the original problem.
It is easy to see that the 10 numbers have exactly 5 even numbers and 1 multiple of 10 at the beginning and end. Therefore, the number of even numbers and the number of multiples of 10 remain unchanged throughout the operations.
Let $a$ be the number among the initial 10 numbers whose unit digit is 5. Consider the first operation involving $a$, and let $b$ be the other number involved.
If $b$ is odd, then $a^{2}-2011 b^{2}$ is even. Thus, the number of even numbers will increase, leading to a contradiction.
If $b$ is even but not a multiple of 10, then $a b$ is a multiple of 10. Thus, the number of multiples of 10 will increase, leading to a contradiction.
If $b$ is a multiple of 10, then the two resulting numbers are both multiples of 25. This means that among the 10 consecutive positive integers obtained at the end, there are two multiples of 25, leading to a contradiction.