Consider a deck of $n > 1$ cards, each with a positive integer written on it. It is given that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards. We need to determine for which values of $n$ it follows that the numbers on the cards are all equal.

### Analyzing the Properties

Let's denote the numbers on the cards as $a_1, a_2, \ldots, a_n$. For any pair of cards $(a_i, a_j)$, the arithmetic mean is given by:

$$\text{Arithmetic Mean} = \frac{a_i + a_j}{2}$$

The condition states that this arithmetic mean equals the geometric mean of some collection of numbers from the cards. Recall that the geometric mean $G(x_1, x_2, \ldots, x_k)$ of a set of numbers is:

$$G(x_1, x_2, \ldots, x_k) = (x_1 \cdot x_2 \cdot \ldots \cdot x_k)^{1/k}$$

### Condition for All Numbers Being Equal

To satisfy the condition, consider the simple case when all numbers on the cards are equal. Let all numbers be $a$. Then for any pair $(a_i, a_j)$:

$$\text{Arithmetic Mean} = \frac{a + a}{2} = a$$

And trivially, the geometric mean of any single card $a$ or any collection of the same numbers will also be $a$. Therefore, if all $a_i$ are equal, the condition is naturally satisfied.

### General Case Verification

For $n = 2$, the given condition must hold for any two numbers $a_1$ and $a_2$:

$$\frac{a_1 + a_2}{2} = \sqrt{a_1 \cdot a_2}$$

Squaring both sides, we get:

$$\left(\frac{a_1 + a_2}{2}\right)^2 = a_1 \cdot a_2 \implies \frac{a_1^2 + 2a_1a_2 + a_2^2}{4} = a_1a_2$$

Simplifying:

$$a_1^2 - 2a_1a_2 + a_2^2 = 0$$

This can be rewritten as:

$$(a_1 - a_2)^2 = 0 \implies a_1 = a_2$$

Thus, for $n = 2$, the numbers must indeed all be equal. Since this condition needs to hold for every pair of cards for $n > 2$, the reasoning extends naturally, as if any two numbers differed, it would violate the established condition.

In conclusion, the property that the numbers must be equal holds for $n$ in general:

$$\boxed{n \text{ for all } n > 1}$$