19. (1991 Asia-Pacific Mathematical Olympiad) During the break, $n(n \geqslant 2)$ students sit in a circle around the teacher to play a game. The teacher distributes candies to the students in a counterclockwise direction according to the following rules: first, the teacher selects a student and gives them a candy, then skips one student to give the next student a candy, then skips 2 students to give the next student a candy, then skips 3 students to give the next student a candy, and so on. Determine the value of $n$ such that, after many rounds (perhaps many circles), each student has at least one candy.
19. The problem is equivalent to determining the positive integer $n$, such that the congruence
$$
1+2+3+\cdots+x=a(\bmod n)
$$

has solutions for $a=0,1,2, \cdots, n-1$ (which is actually equivalent to having solutions for any integer $a$).
Try special values of $n$ and examine whether $1+2+3+\cdots+x$ modulo $n$ can cover a complete residue system modulo $n$.
When $n=2$, $1 \equiv 1(\bmod 2), 1+2 \equiv 1(\bmod 2), 1+2+3 \equiv 0(\bmod 2)$, so $n=2$ satisfies the requirement.
When $n=3$, $1=1(\bmod 3), 1+2 \equiv 0(\bmod 3), 1+2+3 \equiv 0(\bmod 3)$, it is easy to see that as the positive integer $x$ increases, $1+2+3+\cdots+x$ modulo 3 is a periodic sequence with a period of $1,0,0$, so $n=3$ does not satisfy the requirement.

When $n=4$, $1=1(\bmod 4) \cdot 1+2 \equiv 3(\bmod 4), 1+2+3 \equiv 2(\bmod 4), 1+2+3+4 \equiv 2(\bmod 4), 1+2+3+4+5 \equiv 3(\bmod 4), 1+2+3+4+5+6 \equiv 1(\bmod 4), 1+2+3+4+5+6+7 \equiv 0(\bmod 4)$, from which it can be seen that $1+2+3+\cdots+x$ modulo 4 covers a complete residue system $0,1,2,3$, so $n=4$ satisfies the requirement.

When $n=5$, $1 \equiv 1(\bmod 5), 1+2 \equiv 3(\bmod 5), 1+2+3 \equiv 1(\bmod 5), 1+2+3+4 \equiv 0(\bmod 5), 1+2+3+4+5 \equiv 0(\bmod 5)$, the subsequent remainders will inevitably repeat. So $n=5$ does not satisfy the requirement.
Similarly, it can be seen that $n=6,7,9,10$ do not satisfy the requirement, while $n=8$ does satisfy the requirement.
Thus, it is conjectured that $n$ is a power of 2, i.e., $n=2^{k}(k \in \mathbf{N})$, when each of the $n$ students can receive candy.
The next question is: How to prove this conjecture is true?
Re-examine the analysis process of special cases, it is not difficult to find that when $n=3,5,7$, the remainders of $1+2+3+\cdots+x$ modulo $n$ are a very simple cycle. Thus, the following auxiliary conjecture can be made.
Conjecture When $n$ is an odd prime $p$, there must exist an integer $a$ such that (1) has no solution.
The following proves that this auxiliary conjecture is true.
Consider the first $p$ values of $1+2+3+\cdots+x$
$$
1.1+2,1+2+3, \cdots, 1+2+\cdots+(p-1), 1+2+3 \cdots+(p-1)+p,
$$

Since $p$ is an odd prime, we get
$$
1+2+\cdots+(p-1)+p=\frac{p(p+1)}{2} \equiv 0(\bmod p),
$$

so
$$
1+2+\cdots+(p-1) \equiv 1+2+\cdots+(p-1)+p \equiv 0(\bmod p) .
$$

Therefore, the $p$ numbers in (2) have at most $p-1$ different values modulo $p$, and the remainders of $1+2+3+\cdots+x$ modulo $p$ are all periodic, hence there must exist an integer $a$ such that (1) has no solution.
Furthermore, when $n$ has an odd prime factor $p$, if the congruence (1) has solutions for any integer $a$, then the congruence
$$
1+2+3+\cdots+x=a(\bmod p)
$$

must also have solutions for any integer $a$, but this is impossible according to the auxiliary conjecture. Therefore, when $n$ has an odd prime factor $p$, the congruence (1) must have cases with no solution. Thus, to make the congruence (1) have solutions for any integer $a$, $n$ can only be a power of 2.

Next, it needs to be proven that when $n=2^{k}\left(k \in \mathbf{N}_{+}\right)$, the congruence (1) indeed has solutions for any integer $a$. For this, it is only necessary to construct $2^{k}$ integers that are pairwise incongruent modulo $2^{k}$, such that $\frac{x(x+1)}{2}$ can take each of these integers. In fact, the following $2^{k}$ integers meet the requirement (proof omitted).
$$
\frac{0 \times 1}{2}, \frac{1 \times 2}{2}, \frac{2 \times 3}{2}, \cdots, \frac{\left(2^{k}-1\right) \cdot 2^{k}}{2}
$$

Note Sometimes, the conclusion of a proposition is not clear, making it difficult to solve the problem. In such cases, analyzing special cases can lead to relevant conjectures, making the conclusion clear, and then proving it.