Task 3. (Option 2)

Thirty clever students from 6a, 7a, 8a, 9a, and 10a grades were tasked with creating forty problems for the olympiad. Any two classmates came up with the same number of problems, while any two students from different grades came up with a different number of problems. How many people came up with just one problem?
Solution: 26 classmates solved 1 problem, the 27th person solved 2, the 28th solved 3, the 29th solved 4, and the 30th solved 5. This solution is immediately apparent. Let's prove that it cannot be otherwise.

Let $x$ be the number of people who solved one problem, $y$ be the number who solved two, $z$ be the number who solved three, $q$ be the number who solved four, and $r$ be the number who solved five problems. According to the problem, $x, y, z, q, r$ are all at least 1. 
\[
\left\{\begin{array}{c}
x+2 y+3 z+4 q+5 r=40, \\
x+y+z+q+r=30 ;
\end{array}\right.
\]

Subtract the second equation from the first: $y+2 z+3 q+4 r=10$. This equality, given the conditions of our problem, can only be achieved when $y=z=q=r=1$, which means $x=26$.

Answer: 26 people.