8.83 On the blackboard, there are natural numbers $1,2, \cdots, n(n \geqslant 3)$. It is allowed to erase any two numbers $p$ and $q$ each time, and replace them with $p+q$ and $|p-q|$. It is known that after several such modifications, all the numbers on the blackboard are $k$. Find all possible values of $k$.

---

The translation maintains the original format and line breaks as requested.
[Solution] Clearly, all numbers that appear during the operations are non-negative integers. 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 $k$ on the blackboard can be divided by an odd number $d>1$, then all the numbers on the blackboard at the beginning should be multiples of $k$, which is impossible, because 1 cannot be a multiple of $d$. So, $k$ cannot have an odd prime factor, hence $k=2^{s} ; s \in N$. Since the maximum value of the numbers on the blackboard does not decrease after each operation, we know that $k \geqslant n$. Next, we will prove that $k$ can be any $2^{s}, s \in N$ that is not less than $n$.
Because just one operation can achieve
$\left(2^{m}, 2^{m}\right) \longrightarrow\left(0,2^{m+1}\right) ;\left(0,2^{m}\right) \longrightarrow\left(2^{m}, 2^{m}\right)$,

it is not difficult to see that if a set of numbers composed of different powers of 2 is written on the blackboard, and their exponents do not exceed $s_{0}$, and there are two equal numbers less than $2^{s_{0}}$, then after several steps of operation, all the numbers can certainly become $2 \%$.

We use induction on $n \geqslant 3$ to prove: starting from the array $1,2, \cdots, n$, it is always possible to obtain an array composed entirely of powers of 2 after several steps of operation, and $s_{0}$ can be any natural number that satisfies the inequality $2^{s_{0}} \geqslant n$.

Because $(3,1) \rightarrow(4,2)$, the conclusion holds for $n=3,4$. Also, because $(5,3) \longrightarrow$ $(8,2),(6,2) \rightarrow(8,4),(7,1) \rightarrow(8,6)$, the conclusion also holds for $n=5,6,7,8$. Suppose the conclusion holds for all $n$ such that $3 \leqslant n \leqslant m$, we will prove that the conclusion also holds for $n=m+$ $1>8$. Let $m+1=2^{h}+r$, where $h \geqslant 3,1<r \leqslant 2^{h}$. At this time, we perform operations on the pairs $(p, q)=\left(2^{h}+j, 2^{h}-j\right)$, $j=1,2, \cdots, r$, and we can obtain an array composed of the following three parts:
(1) $1,2, \cdots, 2^{h}-r-1$ (these numbers are the numbers on the blackboard that have not been operated on, and this part does not exist when $r=2^{h}$ or $2^{h}-1$);
(2) $2,4, \cdots, 2 r$ (these numbers come from the difference $|p-q|$);
(3) $2^{h}, 2^{h+1}, \cdots, 2^{h+1}$ (these numbers come from the sum $p+q$ and the number $2^{h}$).
Since $h \geqslant 3$, at least one of the first two groups of numbers has at least 3 numbers. When one of the first two groups of numbers has at least 3 numbers, the induction hypothesis can be used. And when one of the groups has at most 2 numbers, they are already all powers of 2. Therefore, starting from $1,2, \cdots, m+1$, after several operations, all the numbers can become powers of 2. On the other hand, when $r=2^{h}-j, j=0,1,2,3$, each of the last two groups has one $2^{h}$. For the remaining $r$, each of the first two groups has one 2. This completes the induction proof.