lasha wrote: $ 30$ students participated in the mathematical Olympiad. Each of them was given $ 8$ problems to solve. Jury estimated their work with the following rule: 1) Each problem was worth $ k$ points, if it wasn't solved by exactly $ k$ students; 2) Each student received the maximum possible points in each problem or got $ 0$ in it; Lasha got the least number of points. What's the maximal number of points he could have? Remark: 1) means that if the problem was solved by exactly $ k$ students, than each of them got $ 30 - k$ points in it. Maximum number of points is $ 60$. Example: if each of first $ 15$ students will solve only first $ 4$ problems, and each of left $ 15$ students will solve onle the last $ 4$ problems, then each of them will receive exactly $ 60$ points. Suppose that $ 60$ points isn't maximum then Lasha got at least $ 61$ points(therefore everyone got at least $ 61$ points) So therefore sum of all students score is at least $ 61 \cdot 30=1830$ points. There is at least one problem such the sum of scores from this problem is $ [\frac{1830}{8}]+1=229$ points. The sum of scores of the students for $ 1$ problem is $ k(30-k)$ Maximum of such sum is $ 225$. Therefore we have contradiction. So we have that maximal points that Lasha could have is $ 60$.

What if we demand Lasha to be the only student with the least number of points? What's the maximal number of points he could then have?

socrates wrote: What if we demand Lasha to be the only student with the least number of points? What's the maximal number of points he could then have?