390. There are 200 candies. What is the minimum number of schoolchildren to whom these candies can be distributed so that among them, there will always be at least two who receive the same number of candies (which could be zero)?
$\triangle$ Let's find the minimum $n$ such that $200400$. There is no need to solve this quadratic inequality in the usual way. Since the product $n(n-1)$ increases with $n$, the minimum $n$ satisfying the inequality can be found by trial and error: $n=21$.

This means that if there are 21 schoolchildren, there will always be two of them who receive the same number of candies.

Let's also verify that with $n=20$, it is possible to distribute the candies such that any two schoolchildren receive a different number of candies. For this, we will give 19 out of the 20 schoolchildren $0,1,2, \ldots, 18$ candies respectively. In total, this is $0+1+2+\ldots+18=\frac{18 \cdot 19}{2}=171$ candies, and thus $200-171=29$ candies remain undistributed. We will give these to the last, twentieth schoolchild.

Answer: the minimum number of schoolchildren is 21.