4. Given an integer $n>0$. There is a balance scale and $n$ weights with masses $2^{0}, 2^{1}, \cdots, 2^{n-1}$.

Now, through $n$ steps, all the weights are to be placed on the balance one by one, ensuring that during the process, the weight on the left side never exceeds the weight on the right side. In each step, one of the weights not yet placed on the balance is chosen and placed on either the left or the right side of the balance, until all weights are placed on the balance. Find the number of different ways to perform this process.
4. (Based on the solution by Yao Bowen)

The number of different methods in the operation process is
$(2 n-1)!!=1 \times 3 \times \cdots \times(2 n-1)$.
We will use mathematical induction on $n$.
When $n=1$, there is only one weight, which can only be placed on the left side of the balance, so there is only 1 method.

Assume that when $n=k$, $k$ weights of $2^{0}, 2^{1}, \cdots, 2^{k-1}$ have $(2 k-1)!!$ methods according to the problem's requirements.

When $n=k+1$, if we multiply all the weights by $\frac{1}{2}$, it does not affect the essence of the problem. At this point, the weights of the $k+1$ weights are $\frac{1}{2}, 1, 2, \cdots, 2^{k-1}$.
Since for any positive integer $r$ we have
$$
2^{r}>2^{r-1}+2^{r-2}+\cdots+1+\frac{1}{2} \geqslant \sum_{i=-1}^{r-1} \pm 2^{i} \text {, }
$$

Thus, when all the weights are placed on the balance, the heavier side of the balance is determined solely by the position of the heaviest weight. Therefore, the heaviest weight must be on the left.

Next, consider the position of the weight $\frac{1}{2}$ during the operation process.
(1) If the weight $\frac{1}{2}$ is placed first, it can only be placed on the left side, and then the remaining $k$ weights have $(2 k-1)!!$ methods.
(2) If the weight $\frac{1}{2}$ is placed at the $t$-th operation ($t=2,3, \cdots, k+1$), since the weights already placed on the balance are all greater than $\frac{1}{2}$, the weight $\frac{1}{2}$ will not be the heaviest. Therefore, it can be placed on either side without affecting the position of the heaviest weight. Hence, there are 2 methods. The remaining weights' placement is not affected, so there are $2 k \times (2 k-1)!!$ methods.
In summary, when $n=k+1$, there are
$$
(2 k-1)!! + 2 k \times (2 k-1)!! = (2 k+1)!!
$$

methods.
By mathematical induction, for any integer $n>0$, the number of different methods in the entire operation process is $(2 n-1)!!$.