11. (20 points) If the subset $A$ of the set $M=\{1,2, \cdots, 200\}$ consists of elements each of which can be expressed as the sum of the squares of two natural numbers (allowing the same number), find the maximum number of elements in the set $A$.
11. Notice that, the squares not exceeding 200 are $0^{2}, 1^{2}, 2^{2}, \cdots, 14^{2}$.
First, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form $k^{2}+0^{2}$, and there are 14 such numbers; while the sum of each pair of numbers (allowing repetition) in $1^{2}$, $2^{2}, \cdots, 10^{2}$ is in the set $M$, and there are $\mathrm{C}_{10}^{2}+10=55$ such numbers, among which, numbers of the form $x^{2}+x^{2}$ are 10, and numbers of the form $x^{2}+y^{2}(x \neq y)$ are $\mathrm{C}_{10}^{2}$.

Second, numbers of the form $11^{2}+x^{2}(x=1,2, \cdots, 8)$ are 8,
numbers of the form $12^{2}+x^{2}(x=1,2, \cdots, 7)$ are 7, numbers of the form $13^{2}+x^{2}(x=1,2, \cdots, 5)$ are 5, and numbers of the form $14^{2}+x^{2}(x=1,2)$ are 2, totaling 22.
Consider the repeated cases.
Notice that, if
$$
\begin{array}{l}
x=a^{2}+b^{2}, y=c^{2}+d^{2}(a \neq b, c \neq d), \\
\text { then } x y=(a c+b d)^{2}+(a d-b c)^{2} \\
=(a c-b d)^{2}+(a d+b c)^{2} .
\end{array}
$$

Numbers not exceeding 40 that can be expressed as the sum of squares of two different positive integers are
$$
5 、 10 、 13 、 17 、 20 、 25 、 26 、 29 、 34 、 37 、 40,
$$

The product of each number in this group with 5, and $13^{2}$, are all in the set $M$, and can each be expressed as the sum of squares in two ways, hence each is counted twice, totaling 12 repetitions (10, $13 、 17 、 20$ multiplied by 10 are already included in the above product group).
Therefore, the maximum number of elements in set $A$ is
$$
14+55+22-12=79 \text {. }
$$