9. (This question is worth 14 points) Find the smallest positive integer $n$, such that for any positive integer $k \geq n$, in the set $M=\{1,2, \cdots, k\}$ of the first $k$ positive integers, for any $x \in M$, there always exists another number $y \in M(y \neq x)$, such that $x+y$ is a perfect square.
It is easy to see that when $n \leq 6$, in $M=\{1,2,3,4,5,6\}$, the number 2 added to any other number is never a square number.
The following proof shows that the minimum value of $n$ is 7.
If positive integers $x, y (x \neq y)$ satisfy: $x+y=$ a square number, then $\{x, y\}$ is called a "square pair". Clearly, in $M=\{1,2, \cdots, 7\}$, $\{1,3\},\{2,7\},\{3,6\},\{4,5\}$ are square pairs; in $M=\{1,2, \cdots, 7,8\}$, the square pair $\{1,8\}$ is added; in $M=\{1,2, \cdots, 7,8,9\}$, the square pair $\{7,9\}$ is added. The following uses induction. A $k$ that satisfies the condition in the problem is said to have property $P$; it is briefly denoted as $k \in P$. According to the above, when $7 \leq k \leq 3^{2}$, all have $k \in P$. Suppose it has been proven that when $7 \leq k \leq m^{2} (m \geq 3)$, all have $k \in P$.

Now consider the case $7 \leq k \leq(m+1)^{2}$. Using the induction hypothesis, it is only necessary to prove that when $k=m^{2}+r$, where $1 \leq r \leq 2 m+1$, all have $k \in P$. First, when $r=2 m+1$, i.e., $k=(m+1)^{2}$, $(k, 2 m+3)$ forms a square pair, because:
$$
k+(2 m+3)=(m+1)^{2}+(2 m+3)=(m+2)^{2}
$$

And by $m^{2}-(2 m+3)=(m-1)^{2}-4 \geq 0$, we know that $2 m+3 \leq m^{2}$, i.e., $2 m+3 \neq k$; when $1 \leq r \leq 2 m$, $(k, 2 m+1-r)$ forms a square pair, because:
$$
k+(2 m+1-r)=\left(m^{2}+r\right)+(2 m+1-r)=(m+1)^{2}
$$

And $1 \leq 2 m+1-r \leq 2 m < m^{2}$, so $2 m+1-r \neq k$.
Therefore, for each $k$ satisfying $7 \leq k \leq(m+1)^{2}$, all have $k \in P$, thus for all positive integers $k$ satisfying $7 \leq k \leq m^{2} (m \geq 3)$, all have $k \in P$, i.e., for all positive integers $k \geq 7$, all have $k \in P$; so the minimum value of $n$ is 7.
(14 points)