8-139 $K$ is a positive integer, $r_{n}$ is the remainder of $C_{2 n}^{n}$ divided by $K$, $0 \leqslant r_{n} \leqslant K$ - 1. Find all $K$ such that the sequence
$$
r_{1}, r_{2}, r_{3}, \cdots
$$

is periodic for all $n \geqslant p$, where $p$ is a fixed positive integer (i.e., it is periodic from some term onwards).
[Solution] When $K=1$, for all $n$ we have
$$
C_{2 n}^{n} \equiv 0(\bmod K)
$$

Thus, for all $n$, $r_{n}=1$. Therefore, $K=1$ is one of the solutions.
When $k=2$, for all $n$ we have
$C_{2 n}^{n}=C_{2 n-1}^{n}+C_{2 n-1}^{n-1}=2 \cdot C_{2 n-1}^{n-1} \equiv 0(\bmod 2)$, similarly, $K=2$ is also one of the solutions.
Next, we discuss the case when $K=4$.
For this, we first prove: when $x$ is an integer and $m$ is a positive integer,
$$
(1+x)^{2^{m}} \equiv 1+2 x^{2^{m-1}}+x^{2^{m}}(\bmod 4) .
$$

Using mathematical induction.
When $m=1$, (1) is an identity.
Assume for some positive integer $m$, (1) holds.
Consider the case for $m+1$, we have
$$
\begin{aligned}
& (1+x)^{2^{m+1}}=\left((1+x)^{2^{m}}\right)^{2} \\
\equiv & \left(1+2 x^{2^{m-1}}+x^{2^{m}}\right)^{2}(\bmod 4) \\
\equiv & 1+x^{2^{m+1}}+2 x^{2^{m}}(\bmod 4) .
\end{aligned}
$$

By the principle of mathematical induction, (1) is proven.
When $n=2^{j}$ (where $j$ is a non-negative integer),
$$
(1+x)^{2 n}=(1+x)^{2^{j+1}},
$$

The coefficient of $x^{n}$ (i.e., $x^{2^{j}}$) in its expansion is $C_{2 n}^{n}$. In (1), let $m=j+1$, we get
$$
(1+x)^{2^{j+1}} \equiv 1+2 \cdot x^{2^{j}}+x^{2^{j+1}}(\bmod 4) .
$$

Therefore,
$$
C_{2 n}^{n} \equiv 2(\bmod 4) \text {. }
$$

When $n \neq 2^{j}$ (where $j$ is a non-negative integer), and $n \geqslant 3$, let
$$
n=2^{j}+r, 0<r<2^{j} \text {. }
$$

By (1),
$$
\begin{array}{c}
(1+x)^{2 n}=(1+x)^{2^{j+1}} \cdot(1+x)^{2 r} \\
\equiv\left(1+2 x^{2^{j}}+x^{2^{j+1}}\right)(1+x)^{2 r}(\bmod 4) .
\end{array}
$$

The coefficient of $x^{n}$ in the left side is $C_{2 n}^{n}$, and in the right side, noting that
$$
2 r<n<2^{j+1},
$$

Therefore, the coefficient of $x^{n}$ in $\bmod 4$ is $2 \cdot C_{2 r}^{r}$. Thus we have
$$
C_{2 n}^{n} \equiv 2 C_{2 r}^{r} \equiv 0(\bmod 4),
$$

Here we used the previously proven result: $C_{2 r}^{r} \equiv 2(\bmod 4)$.
Since when $n=2^{j}$ (where $j$ is a non-negative integer), $C_{2 n}^{n} \equiv 2(\bmod 4)$; and when $n=2^{j}+$ $r\left(j\right.$ is a non-negative integer, $\left.0<r<2^{j}\right)$, $C_{2 n}^{n} \equiv 0(\bmod 4)$, therefore, the sequence $\left\{C_{2 n}^{n} \mid n\right.$ $\in N\}$ is not a periodic sequence in $\bmod 4$, $K=4$ does not meet the requirements of the problem.
Next, we discuss the case when $k=p$ (where $p$ is an odd prime).
Since $p$ is an odd prime, $C_{p}^{i}(i=1,2, \cdots, p-1)$ is a multiple of $p$, so for any integer $x$, we have
$$
(1+x)^{p} \equiv 1+x^{p}(\bmod p) .
$$

Assume for some positive integer $m$, we have
$$
(1+x)^{p^{m}} \equiv 1+x^{p^{m}}(\bmod p) .
$$

Then, we have
$$
(1+x)^{p^{m+1}}=\left((1+x)^{p^{m}}\right)^{p}
$$
$$
\begin{array}{l}
\equiv\left(1+x^{p^{\prime \prime}}\right)^{p}(\bmod p) \\
\equiv 1+\left(x^{x^{\prime \prime}}\right)^{p}(\bmod p) \\
=1+x^{p^{m+1}} .
\end{array}
$$

By the principle of mathematical induction, for any positive integer $m$ and any integer $x$, we have
$$
(1+x)^{p^{\prime \prime}} \equiv 1+x^{p^{\prime \prime \prime}}(\bmod p) .
$$

Using this conclusion, for $j=1,2, \cdots, p^{m}-1$, we have
$$
\begin{array}{l}
(1+x)^{p^{m}+j}=(1+x)^{p^{m}} \cdot(1+x)^{j} \\
\equiv\left(1+x^{p^{\prime \prime \prime}}\right)(1+x)^{j}(\bmod p), \\
\end{array}
$$

Clearly, when $2 n=p^{m}+j\left(j=1,3, \cdots, p^{m}-1\right)$, since $j<n<p^{m}$, we have $C_{2 n}^{n} \equiv 0(\bmod p)$.
When $2 n=2 p^{m}$, since
$$
\begin{aligned}
& (1+x)^{2 p^{m}}=\left((1+x)^{p^{m}}\right)^{2} \\
\equiv & \left(1+x^{p^{m}}\right)^{2}(\bmod p) \\
= & 1+2 x^{p^{m}}+x^{2 p^{m}} .
\end{aligned}
$$

Thus, $C_{2 n}^{\prime \prime}=C_{2 p}^{p_{p}^{\prime \prime m}} \equiv 2(\bmod p)$.
Since
$$
C_{2 n}^{n} \equiv\left\{\begin{array}{l}
0(\bmod p) ， \text { when } 2 n=p^{m}+j\left(j=1,3, \cdots, p^{m}-1\right) \text {, } \\
2(\bmod p), \text { when } 2 n=2 p^{m} \text {, }
\end{array}\right.
$$

Therefore, $\left\{C_{2 n}^{n} \mid n \in N\right\}$ is not a periodic sequence in $\bmod p$, $k=p(p$ is an odd prime $)$ does not meet the requirements of the problem.
The remaining $K$ are all composite numbers, and they belong to at least one of the following two cases:
(1) $K=4 l$ (positive integer $l \geqslant 2$ ),
(2) $K=p l(p$ is an odd prime, positive integer $l \geqslant 2)$.
If the sequence $\left\{C_{2 n}^{n} \mid n \in N\right\}$ modulo the above $k$ is periodic from a certain term, then the sequence modulo 4 or $p$ is also periodic from that term. But this has been proven impossible.
In summary, the only positive integers $k$ that satisfy the problem are 1 and 2.