18. Let the sequence be
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots
$$

Question: (1) What is the value of the 2010th term in this sequence?
(2) In this sequence, what is the index of the 2010th term that has a value of 1?
Three, 18. (1) Group the sequence:
$$
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right), \cdots,\left(\frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}\right), \cdots
$$

Notice that
$$
\begin{array}{l}
1+2+\cdots+62=1953, \\
1+2+\cdots+63=2016 .
\end{array}
$$

Thus, the 2010th term of the sequence belongs to the 63rd group, and is the 7th number from the end of this group, which is $\frac{57}{7}$.
(2) From the grouping in (1), we know that each odd group contains one 1. Therefore, the 2010th 1 appears in the 4019th group, and the 1 in the 4019th group is located at the 2010th position in that group. Hence, the sequence number of the 2010th term with a value of 1 is
$$
(1+2+\cdots+4018)+2010=8076181 .
$$