Problem 4. All students in the class scored different numbers of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did the winner score?
Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.

Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise, $a_{1}+a_{2}+a_{3} \leq 6+7+8=21$).

Therefore, between $a_{3}$ and $a_{n-2}$, there can only be the numbers $10,11,12,13,14$. Their sum is 60, but it should be 47. Therefore, all these numbers are present except for the number 13.

Thus, we get the sequence $a_{1}, a_{2}, 9,10,11,12,14,15, a_{n-1}, a_{n}$, where $a_{1}+a_{2}=14$ and $a_{n-1}+a_{n}=34$. Therefore, $n=10$, $a_{1}=6$, $a_{2}=8$, $a_{n-1}=16$, and $a_{n}=18$.

Answer: A) 10 students; B) 18 points.