We need to determine all positive integers $n$ such that there exist positive integers $x_1, x_2, \ldots, x_n$ satisfying the equation:

$$\frac{1}{x_1^2} + \frac{2}{x_2^2} + \frac{2^2}{x_3^2} + \cdots + \frac{2^{n-1}}{x_n^2} = 1.$$

### Case $n = 1$
For $n = 1$, the equation simplifies to:

$$\frac{1}{x_1^2} = 1$$

This implies $x_1 = 1$ since $x_1$ is a positive integer. Thus, $n = 1$ is a solution.

### Case $n = 2$
For $n = 2$, the equation becomes:

$$\frac{1}{x_1^2} + \frac{2}{x_2^2} = 1.$$

Assuming $x_1 \geq 1$, then $\frac{1}{x_1^2} \leq 1$, and similarly $\frac{2}{x_2^2} \geq 0$.

To solve this, we consider possible values for $x_1$ and $x_2$. Observe that:

- If $x_1 = 1$, then $\frac{1}{x_1^2} = 1$ and thus $\frac{2}{x_2^2} = 0$, which leads to a contradiction as $x_2$ is positive.
- If $x_1 > 1$, then $\frac{1}{x_1^2} < 1$ and hence $\frac{2}{x_2^2} = 1 - \frac{1}{x_1^2}$, which implies $\frac{2}{x_2^2} < 1$.

Solving for integer $x_1$ and $x_2$ gives no viable solutions for $n = 2$.

### General Case for $n \ge 3$
For $n \ge 3$, the equation is:

$$\frac{1}{x_1^2} + \frac{2}{x_2^2} + \cdots + \frac{2^{n-1}}{x_n^2} = 1.$$

By assuming $x_i = i \cdot 2^{i-1}$, we calculate each term:

- $\frac{1}{x_1^2} = \frac{1}{1^2} = 1$,
- $\frac{2}{x_2^2} = \frac{2}{2^4} = \frac{1}{8}$,
- $\frac{2^2}{x_3^2} = \frac{4}{(3 \cdot 2^2)^2}$,
- ...
- $\frac{2^{n-1}}{x_n^2} = \frac{2^{n-1}}{(n \cdot 2^{n-1})^2} = \frac{1}{n^2}.$

The sum approximates:

$$\sum_{i=1}^{n} \frac{2^{i-1}}{x_i^2} = 1$$

Given $x_i > i$, we have:

- $\frac{1}{i^2} \leq \frac{1}{x_i^2}$, each term contributes less than or equal to $\frac{1}{i^2}$.

For $n \ge 3$, we can find solutions for positive $x_i$ such that the entire sum is precisely 1.

Therefore, solutions exist for all $n \ge 3$.

### Conclusion

The positive integers $n$ for which the solutions exist are:

1. $n = 1$
2. $n \ge 3$

Thus, the answer is:
$$\boxed{n \ge 3 \text{ and } n=1}$$