Three. (25 points) If a number $a$ can be expressed as the sum of the squares of two natural numbers (allowing the same), then $a$ is called a "good number". Determine how many good numbers there are among the first 200 positive integers $1,2, \cdots, 200$.

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Three, the squares not exceeding 200 are $0^{2}, 1^{2}, \cdots, 14^{2}$.
Obviously, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form of $k^{2}+$ $0^{2}$, and there are 14 such numbers.

Each pair of numbers in $1^{2}, 2^{2}, \cdots, 10^{2}$ (which can be the same) has a sum not exceeding 200, and there are $\frac{10 \times 9}{2}+10=55$ such numbers (among which, numbers in the form of $x^{2}+x^{2}$ are 10, and numbers in the form of $x^{2}+y^{2}(x \neq y)$ are $\frac{10 \times 9}{2}=45$).

Next, numbers in the form of $11^{2}+x^{2}(x=1,2, \cdots, 8)$ are 8; numbers in the form of $12^{2}+x^{2}(x=1,2, \cdots, 7)$ are 7; numbers in the form of $13^{2}+x^{2}(x=1,2, \cdots, 5)$ are 5; and numbers in the form of $14^{2}+x^{2}(x=1,2)$ are 2, totaling 22.

Considering the repeated cases, use the following fact:
$$
\begin{array}{l}
\text { If } x=a^{2}+b^{2}, y=c^{2}+d^{2}(a \neq b, c \neq d), \text { then } \\
x y=(a c+b d)^{2}+(a d-b c)^{2} \\
=(a c-b d)^{2}+(a d+b c)^{2} .
\end{array}
$$

Numbers not exceeding 40 and can be expressed as the sum of squares of two different positive integers are $5,10,13,17,20,25,26,29,34,37,40$. Each number in this group, when multiplied by 5, and $13^{2}$, does not exceed 200 and can be expressed as the sum of squares in two ways, so each is counted twice, totaling 12 repeated counts $(10,13,17,20$ when multiplied by 10 are already included in the above product group).
Therefore, the total number of numbers that meet the condition is $14+55+22-12=79$.
(Supplied by Tao Pingsheng)