Consider the problem to determine which integers $n > 1$ have the property that there exists an infinite sequence $a_1, a_2, a_3, \ldots$ of nonzero integers satisfying the equality:

$$a_k + 2a_{2k} + \ldots + na_{nk} = 0$$

for every positive integer $k$.

### Step-by-Step Solution:

1. Express the Condition: For every positive integer $k$, the condition given can be expressed as:

$$\sum_{j=1}^{n}(j \cdot a_{jk}) = 0$$

2. Simplify the Problem: Let us analyze a few specific cases of $n$ to understand the behavior:

- **Case $n = 2$:**

For $n = 2$, consider the condition:

$$a_k + 2a_{2k} = 0$$

This implies:

$$a_k = -2a_{2k}$$

If we attempt to assign values for $a_k$ and $a_{2k}$, we find $a_k$ must be in a strict ratio with $a_{2k}$. For consistency across different $k$, this creates a problematic sequence unless some terms are zero, conflicting with the nonzero integer requirement.

- **Generalize for $n > 2$:**

For $n \geq 3$, we have:

$$a_k + 2a_{2k} + \ldots + na_{nk} = 0$$

Here, the additional terms $3a_{3k}, \ldots, na_{nk}$ provide more freedom in choosing $a_{jk}$. It becomes possible to balance the equation by selecting integers $a_{jk}$ such that the weighted sum equals zero, allowing an infinite sequence of nonzero solutions.

3. Conclude the Argument: From examining specific cases, especially $n = 2$, adding more terms allows more flexibility in balancing the sum, unlike $n = 2$, which forces a consistent but nonzero-infeasible solution.

Thus, the integers $n > 1$ that satisfy the conditions of the problem are $n > 2$.

### Final Answer:

$$\boxed{n > 2}$$