There are$n$ balls numbered $1,2,\cdots,n$, respectively. They are painted with $4$ colours, red, yellow, blue, and green, according to the following rules:
First, randomly line them on a circle.
Then let any three clockwise consecutive balls numbered $i, j, k$, in order.
1) If $i>j>k$, then the ball $j$ is painted in red;
2) If $i<j<k$, then the ball $j$ is painted in yellow;
3) If $i<j, k<j$, then the ball $j$ is painted in blue;
4) If $i>j, k>j$, then the ball $j$ is painted in green.
And now each permutation of the balls determine a painting method.
We call two painting methods distinct, if there exists a ball, which is painted with two different colours in that two methods.

Find out the number of all distinct painting methods.
To solve this problem, we need to determine the number of distinct painting methods for $n$ balls arranged in a circle and painted according to the given rules. The solution involves using the properties of Catalan numbers. Let's break down the solution step-by-step.

1. **Base Case:**
   For $n = 3$, we can manually verify the distinct painting methods. There are $3! = 6$ permutations of the balls, and each permutation will result in a unique painting method. Thus, the base case holds true.

2. **Inductive Hypothesis:**
   Assume that for $n$ balls, the number of distinct painting methods is given by the $n$-th Catalan number $C_n = \frac{1}{n+1} \binom{2n}{n}$.

3. **Inductive Step:**
   We need to show that for $n+1$ balls, the number of distinct painting methods is given by the $(n+1)$-th Catalan number $C_{n+1}$.

   - Consider the ball numbered $n+1$. This ball can be painted blue if it is the largest number in any three consecutive balls. We need to count the number of ways to arrange the remaining $n$ balls such that the painting rules are satisfied.

   - If the ball $n+1$ is blue, then the remaining $n$ balls can be divided into two groups: one group of $t$ balls (where $2 \le t \le n-1$) and another group of $n-t$ balls. The balls in each group can be arranged in such a way that the painting rules are satisfied.

   - For each $t$, the number of ways to arrange the $t$ balls and the $n-t$ balls is given by the product of two Catalan numbers: $C_t \cdot C_{n-t}$.

   - Summing over all possible values of $t$, we get:
     \[
     C_{n+1} = \sum_{t=1}^{n} C_{t-1} \cdot C_{n-t}
     \]

4. **Conclusion:**
   By the definition of Catalan numbers, we have:
   \[
   C_{n+1} = \sum_{t=0}^{n} C_t \cdot C_{n-t}
   \]
   This matches the recursive formula for Catalan numbers, confirming that the number of distinct painting methods for $n+1$ balls is indeed $C_{n+1}$.

The final answer is $\boxed{\frac{\binom{2n-2}{n-1}}{n}}$