$2 \cdot 24$ Given 2000 points on a circle, one of them is marked with the number 1. Starting from this point, move clockwise to the second point and mark it with the number 2. From the point marked with the number 2, move clockwise to the third point and mark it with the number 3 (as shown in the figure), and continue this process until $1, 2, 3, \cdots, 1993$ are all marked on the points. Some of the points on the circle will be marked with more than one number, and some points will not be marked with any number. What is the smallest integer marked on the point that is marked with 1993?
（11th American Invitational Mathematics Examination, 1993）
[Solution] Label 2000 points in a clockwise direction as $1, 2, 3, \cdots, 2000$.
According to the problem, the labeled numbers are $a_{1}, a_{2}, \cdots, a_{1993}$.
Thus, $a_{1}=1, \quad a_{2}=3, \quad a_{3}=6, \cdots$.
$$
\begin{array}{l}
a_{n}=1+2+\cdots+n=\frac{n(n+1)}{2} . \\
a_{3}=\frac{1993 \cdot 1994}{2}=1987021=1021 \quad(\bmod 2000) .
\end{array}
$$

Therefore, $a_{1993}$ is located at the 1021st position.
Next, find the smallest positive integer $i$ such that
$$
a_{i}=\frac{i(i+1)}{2} \equiv 1021 \quad(\bmod 2000) .
$$

That is, find the smallest positive integer $k$ such that
$$
\begin{array}{c}
\frac{i(i+1)}{2}=1021+2000 k \\
i^{2}+i-2042-4000 k=0 \\
i=\frac{-1 \pm \sqrt{16000 k+8169}}{2}
\end{array}
$$

The smallest $k$ that makes $i$ the smallest positive integer is $k=3$.
At this time, $i=118$, meaning the smallest positive integer marked on the point labeled 1993 (the 1021st position) is 118.