14. A positive integer that can be expressed as the difference of squares of two positive integers is called a "wise number". For example, $9=5^{2}-4^{2}$, so $9$ is a wise number.
(1) Determine which numbers among the positive integers are wise numbers, and explain the reason;
(2) In the sequence of wise numbers arranged in ascending order, find the 2015th wise number.
14. (1) It is easy to know that the positive integer 1 cannot be expressed as the difference of squares of two positive integers, i.e., 1 is not a wise number.
For odd numbers greater than 1, we have
$$
2 k+1=(k+1)^{2}-k^{2}(k=1,2, \cdots),
$$

which means that all odd numbers greater than 1 are wise numbers.
When $k=2,3, \cdots$, we have
$4 k=(k+1)^{2}-(k-1)^{2}$.
When $k=1$, if $4=a^{2}-b^{2}$, then
$$
4=(a+b)(a-b),
$$

we know that there do not exist positive integers $a, b$ that satisfy this equation, i.e., 4 is not a wise number.

Therefore, numbers greater than 4 and divisible by 4 are all wise numbers.

For positive integers of the form $4 k+2(k=0$, $1, \cdots)$, assume $4 k+2$ is a wise number. Then
$$
4 k+2=x^{2}-y^{2}=(x+y)(x-y),
$$

where $x, y$ are positive integers.
When $x, y$ have the same parity, $4 \mid(x+y)(x-y)$, but $4 \nmid(4 k+2)$, which is a contradiction;

When $x, y$ have different parities, $(x+y)(x-y)$ is odd, but $4 k+2$ is even, which is a contradiction.

Therefore, numbers of the form $4 k+2(k=1,2, \cdots)$ are not wise numbers.

Hence, only numbers of the form $2 k+1, 4(k+1)(k=1,2, \cdots)$ are wise numbers.
(2) Among the positive integers starting from 1, every four consecutive numbers form a group. From (1), we know that in the first group, only 3 is a wise number, and in each subsequent group, three out of the four numbers are wise numbers.

Notice that, $2016=3 \times 672$, and $4 \times 672=$ 2688, and since the first group has only 1 wise number, 2688 is the 2014th wise number.
Therefore, 2689 is the 2015th wise number.