Consider an integer $n > 0$ and a balance with $n$ weights of weights $2^0, 2^1, \ldots, 2^{n-1}$. Our task is to place each of these weights on the balance, one by one, so that the right pan is never heavier than the left pan. We aim to determine the number of ways to achieve this.

### Understanding the Problem

The setup involves choosing at each step one of the $n$ weights that has not yet been placed on the balance and deciding whether it should be placed on the left pan or the right pan. This continues until all weights are placed.

### Key Constraints

1. At every step during the placement of the weights, the total weight in the right pan must not exceed the total weight in the left pan.
2. Once a weight is placed, it cannot be moved again.

### Solution Approach

To solve this problem, we consider each weight $2^k$ as a decision point: each weight can either be placed on the left or right pan, constrained by the requirement that the right pan never becomes heavier.

#### Combinatorial Enumeration using Catalan Paths

This is combinatorially equivalent to finding the number of ways to arrange the sequence of weights, where each step of adding a weight to the left is analogous to taking an upward step ($U$), and adding a weight to the right is analogous to a downward step ($D$). For the configuration to satisfy the condition (i.e., right pan never heavier than the left), it is essentially a "path" problem where paths never fall below the "starting level".

The number of distinct configurations achievable with these constraints is closely related to Catalan numbers, calculated in terms of "factorial double" or semifactorials, which specifically articulate the number of valid parenthesis combinations for a sequence of terms.

#### Calculation

The correct formula for the number of valid sequences like described above where the sequence never "falls below ground" is given by the formula:

$$(2n-1)!!$$

where $(2n-1)!!$ denotes the product of all odd integers up to $2n-1$.

Thus, the number of ways the weights can be placed on the balance so that the right pan is never heavier than the left pan is:

$$\boxed{(2n-1)!!}$$

This result reflects the combinatorial counting of valid balanced arrangements, revealing the complexity and richness of the constraining arrangement task akin to classic path and matching problems in combinatorics.