II. (40 points) A class has 25 students. The teacher wants to prepare $N$ candies to host a competition, and distribute the candies based on the scores (students with the same score receive the same number of candies, and students with lower scores receive fewer candies, which can be 0). Question: For what minimum $N$, can the candies be distributed in such a way that no matter how many questions there are in the competition, and regardless of how the students answer the questions?
II. The minimum value of $N$ is 600.
Notice that, if 25 people have the same score, each person must receive the same number of candies, i.e., $25 \mid N$. Let $N=25k$.

Now consider the case where 1 person gets 0 points and 24 people get 1 point.
By the condition, there exist $x>y \geqslant 0 (x, y \in \mathbf{Z})$ such that
$$
24x + y = N = 25k.
$$

By the principle of the average value, we have
$$
x > k > y \Rightarrow x \geqslant k + 1.
$$

Thus, $k - y = 24(x - k) \geqslant 24$
$\Rightarrow k \geqslant 24 + y \geqslant 24$.
Therefore, $N = 25k \geqslant 600$.
We will now prove that when $N = 600$, it is always possible to distribute the candies reasonably.

We prove by induction on $n$: For any positive integer $n$, $n$ students and $n(n-1)$ candies can always be distributed reasonably, and each person gets fewer than $2n$ candies.
When $n=1$, it is obviously true.
Assume the proposition holds for all values less than $n$.
Now consider the case for $n$.
Suppose there are $n-k (0 \leqslant k \leqslant n-1)$ people with the highest score. If $k=0$, then each person gets $n-1$ candies, and the proposition holds. If $1 \leqslant k \leqslant n-1$, by the induction hypothesis, $k(k-1)$ candies can be reasonably distributed to the $k$ people who are not the highest scorers, and each person gets fewer than $2k$ candies. There are
$$
n(n-1) - k(k-1) = (n-k)(n+k-1)
$$

candies left, which are given to the $n-k$ highest scorers, with each receiving $n+k-1$ candies.
Notice that, $2k \leqslant n+k-1 < 2n$.
Therefore, this is a reasonable distribution method, and each person gets fewer than $2n$ candies.