1・185 On the blackboard, there are natural numbers $1,2,3, \cdots, n$, where $n \geqslant 3$. Each time, it is allowed to erase any two numbers $p$ and $q$, and replace them with $p+q$ and $|p-q|$. After such several rewrites, all the numbers on the blackboard become $k$. What values can $k$ possibly be?
[Solution] Let $s$ be any natural number satisfying the inequality $2^{s} \geqslant n$.
Obviously, after each step, non-negative integers will be written on the blackboard. If the sum and difference of two non-negative integers can both be divided by an odd number $d$, then these two numbers themselves can also be divided by $d$. Therefore, if the number $k$ can be divided by an odd number $d>1$, then the numbers initially written on the blackboard should all be divisible by $d$. But this is impossible, because there is the number 1. Therefore, $k$ cannot have odd prime factors, that is, $k=2^{s}$. Since after each step, the largest number on the blackboard does not decrease, it follows that $k \geqslant n$.
Next, we will prove that $k$ can be any $2^{s}$ not less than $n$.
First, because from the pair $\left(2^{a}, 2^{a}\right)$ we can get $\left(0,2^{a+1}\right)$, and from $\left(0,2^{c}\right)$ we can get $\left(2^{c}, 2^{c}\right)$, so if a set of numbers written on the blackboard consists of powers of 2, with their exponents not exceeding $s_{0}$, and there are two equal numbers less than $2^{s_{0}}$, then after a few steps, all the numbers can be made to become $2^{s_{0}}$, and $s_{0}$ can be any natural number satisfying the inequality $2^{s_{0}} \geqslant n$.

We will now prove: starting from the array $1,2, \cdots, n$, it is always possible to obtain the above array consisting entirely of powers of 2.

In fact, when $n=3$, it is obvious: from $1,2,3$ we can get $2,2,4$. Assuming the statement is true for all $n$ satisfying the inequality $3 \leqslant n \leqslant m$, we need to prove that the statement is also true for $n=m+1$.

If $m+1 \leqslant 8$, then it can be directly verified. Now let $m+1=2^{l}+b$, where $1 \leqslant b \leqslant 2^{l}, l \geqslant 3$. We will perform the transformation described in the problem on the numbers $p=2^{l}+u, q=2^{l}-u$, where $1 \leqslant u \leqslant b$ (when $b=2^{l}$, $1 \leqslant u \leqslant b-1$), and we can obtain an array composed of the following three parts:
(1) $1,2, \cdots, 2^{l}-b-1$ (when $b=2^{l}$ and $b=2^{l}-1$, this part does not exist);
(2) $2,4, \cdots, 2 b$ (this part is derived from the difference $|p-q|$);
(3) $2^{l}, 2^{l+1}, \cdots, 2^{l+b}$ (this part is derived from the sum $p+q$ and the number $2^{l}$). If the numbers in the first and second parts are both at least 3 $\left(3 \leqslant b \leqslant 2^{l}-4\right)$, then we can apply the induction hypothesis to each of them. If only one of these parts has at least 3 numbers, then we can apply the induction hypothesis to that part, and the numbers in the other two parts are already powers of 2. If the numbers in the first and second parts are both less than 3, then from $2^{l}-b-1 \leqslant 2$ and $2 b \leqslant 4$ we get $2^{l} \leqslant 5$, which contradicts $l \geqslant 3$, thus the proposition is proved.