7.41 A natural number that can be expressed as the difference of the squares of two natural numbers is called a "wise number". For example, $16=5^{2}-3^{2}$, so 16 is a "wise number". In the sequence of natural numbers, starting from 1, what is the 1990th "wise number"? Please explain your reasoning.
(China Beijing High School Grade 1 Mathematics Competition, 1990)
[Solution] (1) 1 cannot be expressed as the difference of squares of two natural numbers, so 1 is not a "wise number";
(2) For odd natural numbers greater than 1, $2 k+1$, we have
$$
2 k+1=(k+1)^{2}-k^{2} .
$$

Therefore, all odd natural numbers greater than 1 are "wise numbers";
(3) Since $x^{2}-y^{2}=(x+y)(x-y)$, and $x+y$ and $x-y$ have the same parity, the product $(x+y)(x-y)$ is either odd or a multiple of 4. Therefore, numbers of the form $4 k+2$ are not "wise numbers";
(4) For natural numbers of the form $4 k(k>1)$, we have
$$
4 k=(k+1)^{2}-(k-1)^{2} \text {. }
$$

Therefore, natural numbers of the form $4 k(k>1)$ are "wise numbers",
but when $k=1$, $4 k=4$ is not a "wise number".
Thus, among the numbers 1, 2, 3, 4, only 3 is a "wise number", and in each subsequent group of numbers: $4 k+$ $1, 4 k+2, 4 k+3, 4 k+4(k=1,2, \cdots)$, there are 3 "wise numbers".

Since $1989=3 \cdot 663$, the 1990th "wise number" should be produced after the $4 \cdot 663=2652$th number.

And 2652 is the 1987th "wise number", 2653 is the 1988th "wise number", 2654 is not a "wise number", 2655 is the 1989th "wise number", 2656 is the 1990th "wise number".