Initially 17. M $1,2,3,4, \cdots, 1994$ are 1994 numbers from which $k$ numbers are arbitrarily selected, such that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.

In the 1994 numbers $1,2,3,4, \cdots, 1994$, arbitrarily select $k$ numbers, so that any two numbers selected as side lengths uniquely determine a right-angled triangle. Try to find the maximum value of $k$.
Solve for the maximum value of $k$ being 11.
(1) From $A=\{1,2,3,4, \cdots, 1994\}$, select $\{1,2,4,8,16,32,64,128,256,512,1024\}$, a total of 11 numbers, using the recursive formula: $a_{1}=1, a_{n+1}=2 a_{n}$. Any two numbers $a_{i}, a_{j}$ $(i<j)$ in this set satisfy $a_{i}<a_{j}$, and $2 a_{i} \leqslant a_{j}$. Therefore, any two numbers $a_{i}, a_{j}$ can uniquely determine an isosceles triangle. Hence, $k \geqslant 11$.
(2) Suppose $a_{1}<a_{2}<a_{3}<\cdots<a_{12}$ are any 12 numbers selected from $A$, and any two of these numbers can uniquely determine an isosceles triangle. Then, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{3} \geqslant 4, a_{4} \geqslant 8, \cdots, a_{11} \geqslant 1024, a_{12} \geqslant 2048$. This contradicts $a_{12} \leqslant 1994$. This indicates that at most 11 numbers can be selected from $A$ such that any two of them can uniquely determine an isosceles triangle. Therefore, the maximum value of $k$ is 11.
(Tongtan Town, Huxian County, Sichuan, 646111, Answered by Zhong Dingji)