Given a sequence of numbers:
$$
\begin{array}{l}
1, \frac{1}{2}, \frac{2}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \frac{2}{3}, \frac{1}{3}, \\
\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{3}{4}, \frac{2}{4}, \frac{1}{4}, \cdots
\end{array}
$$

Question: (1) What is the position of $\frac{5}{8}$ in the sequence?
(2) What is the 200th number in the sequence?
(1) Observing, we find that there is only 1 number with a denominator of 1, 3 numbers with a denominator of 2, 5 numbers with a denominator of 3, $\cdots \cdots$ and $2n-1$ numbers with a denominator of $n$. Therefore, the numbers with denominators from $1 \sim 7$ total
$$
(1+3+5+\cdots+13)=\frac{14 \times 7}{2}=49 \text { (numbers) }
$$

Also, $\frac{5}{8}$ is the 5th or 11th number in the group of numbers with a denominator of 8, so $\frac{5}{8}$ is the 54th number or the 60th number.
(2) Before the group of numbers with a denominator of $n$, there are $\frac{n(1+2 n-1)}{2}=n^{2}$ numbers. The largest number satisfying $n^{2} \leqslant 200$ is 14, so the 200th number is the 4th number in the group with a denominator of 15, which is $\frac{4}{15}$.