3. $k 4$ ** A natural number that can be expressed as the difference of the squares of two natural numbers is called a "wisdom number". For example, $16=5^{2}-3^{2}$, so 16 is a "wisdom number". In the sequence of natural numbers starting from 1, which is the 1990th "wisdom number"? Please explain your reasoning.
Obviously, 1 is not a "wise number", and odd numbers greater than 1, $2k+1=(k+1)^{2}-k^{2}$, are all "wise numbers".

From $4k=(k+1)^{2}-(k-1)^{2}$, we know that numbers greater than 4 and divisible by 4 are all "wise numbers", but 4 is not a "wise number".

Since $x^{2}-y^{2}=(x+y)(x-y)(x, y \in \mathbf{N})$, when $x$ and $y$ have the same parity, $(x+y)(x-y)$ is divisible by 4. When $x$ and $y$ have different parity, $(x+y)(x-y)$ is odd. Therefore, numbers of the form $4k+2$ are not "wise numbers".

In the sequence of natural numbers, among the first four natural numbers, only 3 is a "wise number". Thereafter, in every consecutive four integers, there are three "wise numbers".
Since $1989=3 \times 663$, $2656=4 \times 664$ is the 1990th "wise number".