Three. (Full marks 25 points) A natural number that can be expressed as the difference of the squares of two natural numbers is called a "smart number." For example, $16=5^{2}-3^{2}$ is a "smart number." Try to answer the following questions:
(1) Is 1998 a "smart number"? Explain your reasoning.
(2) Arranged in ascending order, what is the 1998th "smart number"?
Three, (1) 1998 is not a "smart number". If 1998 were a "smart number", we could set \(1998 = m^2 - n^2\) (where \(m\) and \(n\) are natural numbers), then \((m+n)(m-n) = 2 \times 999\).
Since \(m+n\) and \(m-n\) have the same parity, \((m+n)(m-n)\) is either odd or divisible by 4, hence it cannot be 1998. Therefore, 1998 is not a "smart number".
(2) Let \(k\) be a natural number. Using the method in (1), we can prove that natural numbers of the form \(4k + 2\) are not "smart numbers".
\[
\begin{array}{l}
\text{Also, } 4k = (2k+1)^2 - (2k-1)^2, \\
2k+1 = (k+1)^2 - k^2,
\end{array}
\]
Thus, natural numbers of the form \(4k\) and \(2k+1\) (excluding 1) are all "smart numbers". Arranging "smart numbers" in ascending order:
\[
3, 4, 5, 7, 8, 9, 11, 12, 13, \cdots
\]
Therefore, the \(n\)-th "smart number" is \(n + 2 + \left[\frac{n-1}{3}\right]\) (where \([ \] denotes the floor function). When \(n = 1998\), this "smart number" is
\[
1998 + 2 + \left[\frac{1998-1}{3}\right] = 2665.
\]
(Xu Jianguo, Nanchang Experimental High School, 330006)