We are tasked with finding all positive integers $n$ for which there exist $k \geq 2$ positive rational numbers $a_1, a_2, \ldots, a_k$ satisfying the conditions:

$$a_1 + a_2 + \cdots + a_k = a_1 \cdot a_2 \cdots a_k = n.$$

To find the possible values of $n$, we analyze the problem for small values of $k$:

### Step 1: Case for $k = 2$

Consider $a_1 = a_2 = x$. We then have:

$$2x = x^2 = n.$$

From $2x = x^2$, we get:

$$x^2 - 2x = 0 \implies x(x - 2) = 0.$$

Thus, $x = 2$. Therefore, $n = x^2 = 4$ is one solution with $k = 2$.

### Step 2: Consideration for $k \geq 3$

For $k \geq 3$, let's assume $a_1 = a_2 = \cdots = a_k = x$. Then:

$$kx = x^k = n.$$

This gives us $x = \frac{n}{k}$ and:

$$\left( \frac{n}{k} \right)^k = n.$$

Rearranging gives us:

$$n^{k-1} = k^k.$$

Finding explicit solutions for specific $n$ and $k$:

#### For $n = 6$
Consider $a_1 = a_2 = 3$ and $a_3 = a_4 = \cdots = a_6 = 1$:

$$3 + 3 + 1 + 1 + 1 + 1 = 10 \quad \text{and} \quad 3 \cdot 3 \cdot 1 \cdot 1 \cdot 1 \cdot 1 = 6.$$

This setup doesn't work with pure $k = 3$ identical aspects, but shows $n > k$ can allow different setups like:

- $a_1 = 3$, $a_2 = 2$, $a_3 = 1$.
- This gives $3 + 2 + 1 = 6$ and $3 \cdot 2 \cdot 1 = 6$ which is valid.

Thus, $n = 6$ is also possible.

### General Conclusion

- For $n = 4$, clearly $k = 2$ works.
- For $n \geq 6$, using combinations of $a_1, a_2, \ldots, a_k$ where at least one $a_i > 1$ and other simplifications, solutions exist (like above).

Thus, the possible values of $n$ are:

$$\boxed{4 \text{ or } n \geq 6}.$$