To solve this problem, we need to determine for which integers $n \geq 3$, any set of $n$ positive real numbers $a_1, a_2, \ldots, a_n$, under the condition $\max(a_1, a_2, \ldots, a_n) \leq n \cdot \min(a_1, a_2, \ldots, a_n)$, contains three numbers that can serve as the side lengths of an acute triangle.

### Triangle Inequality Conditions

For three numbers $x, y, z$ to be the side lengths of an acute triangle, they must satisfy:
1. The triangle inequalities:
$$x + y > z, \quad y + z > x, \quad z + x > y.$$
2. The condition for an acute triangle:
$$x^2 + y^2 > z^2, \quad y^2 + z^2 > x^2, \quad z^2 + x^2 > y^2.$$

### Applying the Given Condition

Given the condition $\max(a_1, a_2, \ldots, a_n) \leq n \cdot \min(a_1, a_2, \ldots, a_n)$, denote:
$$M = \max(a_1, a_2, \ldots, a_n), \quad m = \min(a_1, a_2, \ldots, a_n).$$
Then, we have:
$$M \leq n \cdot m.$$

### Establishing an Acute Triangle

We need to check if there exist three numbers among $a_1, a_2, \ldots, a_n$ that can be $x, y,$ and $z$ such that:
$$x^2 + y^2 > z^2.$$

For large $n$, the ratio $\frac{M}{m}$ being small allows the flexibility to select three numbers closer to each other, increasing the likelihood of forming an acute triangle.

### Determining $n$

An example helps illustrate when such sets of numbers are possible, but due to the condition $M \leq n \cdot m$, let's consider when:
$$n \leq 12.$$
In these cases, it is possible to construct examples where three numbers do not satisfy the acute triangle condition, especially if the numbers are spread evenly (such as arithmetic progressions with large gaps).

However, for $n \geq 13$, this constraint becomes less significant, as the maximum spread allowed (i.e., $M/m$) shrinks relative to the number of elements, and we can always find subsets like {1, 1, 1} which trivially satisfy both the triangle inequalities and the acute condition.

Hence, the minimum integer $n$ such that the condition is always satisfied is:
$$n \geq \boxed{13}.$$