60. In a class, there are 25 students. The teacher wants to stock $N$ candies, conduct an olympiad, and distribute all $N$ candies for success in it (students who solve the same number of problems should receive the same number of candies, those who solve fewer should receive fewer, including possibly zero candies). What is the smallest $N$ for which this will be possible regardless of the number of problems in the olympiad and the success of the students?
60. Answer: $600=25 \cdot 24$ candies.

Let's show that a smaller number of candies might not be enough. If all participants solved the same number of problems, the number of candies must be a multiple of 25. Let $N=25 k$. Imagine that 24 people solved the same number of problems, while the 25th solved fewer. If each of the first 24 participants receives $k$ candies or even fewer, there will be extra candies left. Therefore, each of them must receive at least $k+1$ candies, which means $25 k \geqslant 24(k+1)$, or $k \geqslant 24$.

Now we will prove by induction that it is possible to distribute $m$ participants $m(m-1)$ candies such that each receives fewer than $2 s$ candies. The base case for $m=1$ is obvious: give the only participant 0 candies.

The transition from $m-1$ to $m$. Suppose there are $s \geqslant 1$ participants with the highest score, and besides them, there are $t$ participants, $s+t=m$. To the less successful participants, we will distribute $t(t-1)$ candies (each receiving fewer than $2 t$) by the induction hypothesis. After this, $(s+t)(s+t-1)-t(t-1)=s(s+2 t-1)$ candies will remain, and we will distribute them equally among the $s$ "top" participants. Each of them will receive $s+2 t-1$ candies. This number is not less than $2 t$, so it is more than what we distributed to each of the others. Additionally, $s+2 t-1 < 2 s+2 t=2 m$, which completes the transition.