Find all positive integers $n$ with $n \ge 2$ such that the polynomial \[ P(a_1, a_2, ..., a_n) = a_1^n+a_2^n + ... + a_n^n - n a_1 a_2 ... a_n \] in the $n$ variables $a_1$, $a_2$, $\dots$, $a_n$ is irreducible over the real numbers, i.e. it cannot be factored as the product of two nonconstant polynomials with real coefficients.

[i]Proposed by Yang Liu[/i]
To find all positive integers \( n \) with \( n \ge 2 \) such that the polynomial 
\[ P(a_1, a_2, \ldots, a_n) = a_1^n + a_2^n + \cdots + a_n^n - n a_1 a_2 \cdots a_n \]
is irreducible over the real numbers, we will use a combination of algebraic and number-theoretic arguments.

1. **Base Case Analysis:**
   - For \( n = 2 \):
     \[ P(a_1, a_2) = a_1^2 + a_2^2 - 2a_1a_2 \]
     This can be rewritten as:
     \[ P(a_1, a_2) = (a_1 - a_2)^2 \]
     which is clearly reducible.

   - For \( n = 3 \):
     \[ P(a_1, a_2, a_3) = a_1^3 + a_2^3 + a_3^3 - 3a_1a_2a_3 \]
     Using the identity for the sum of cubes:
     \[ a_1^3 + a_2^3 + a_3^3 - 3a_1a_2a_3 = (a_1 + a_2 + a_3)(a_1^2 + a_2^2 + a_3^2 - a_1a_2 - a_2a_3 - a_3a_1) \]
     This shows that \( P(a_1, a_2, a_3) \) is reducible.

2. **General Case for \( n \ge 4 \):**
   - We claim that for \( n \ge 4 \), the polynomial \( P(a_1, a_2, \ldots, a_n) \) is irreducible over the real numbers.
   - Consider the polynomial \( Q(a_1, \ldots, a_n) = a_1^n + \cdots + a_n^n - n a_1 \cdots a_n \).
   - Let \( E = \mathbb{C}(a_1, \ldots, a_n) \) be a field extension. We need to show that \( Q \) is irreducible in \( E \).

3. **Automorphisms and Symmetry:**
   - Let \( \omega \) be a primitive \( n \)-th root of unity. An automorphism \( \sigma \) that fixes \( Q \) must map \( a_k \) to \( \omega^{c_k} a_{\pi(k)} \) for some \( c_k \in \{0, \ldots, n-1\} \) and \( n \mid c_1 + \cdots + c_n \).
   - If \( R \) is an irreducible polynomial in \( E \), then \( \sigma R \) is also irreducible in \( E \).

4. **Homogeneity and Degree Considerations:**
   - All irreducible factors of \( Q \) are homogeneous. Let \( R(a_1, \ldots, a_n) \) be an irreducible factor of minimal degree \( d \). Clearly, \( d \le \frac{n}{2} \).
   - For indices \( j, k \), let \( S_{j,k} = \{ u - v \mid R \text{ has a term with } a_j^u a_k^v \} \). Let \( d_{j,k} \) be the gcd of \( n \) and differences of pairs of elements in \( S_{j,k} \).

5. **Automorphisms and Irreducibility:**
   - There are at least \( \frac{n}{d_{j,k}} \) automorphisms that fix \( Q \) but change \( R \). Assume that all terms in \( R \) are multiplied by \( \omega^t \) for some \( t \). This implies \( c \) is divisible by \( \frac{n}{d_{j,k}} \), so \( c = 0 \) is forced.
   - This produces \( \frac{n}{d_{j,k}} \) distinct irreducible polynomials that are factors of \( Q \).

6. **Conclusion:**
   - Since \( \deg R \le \frac{n}{2} \), there exist indices \( u, v \) such that a term of \( R \) does not include \( a_u \) or \( a_v \). Then \( d_{u,v} \le d \) and shifting gives \( \frac{n}{d} \) distinct irreducible polynomials that divide \( Q \).
   - The product of these polynomials has degree \( n = \deg Q \). Equality holds only if \( d \mid n \) and \( d_{u,v} = d \), implying \( R \) is a polynomial in \( a_1^d, \ldots, a_n^d \). The \( a_1 \cdots a_n \) term in \( Q \) implies \( d = 1 \).

Therefore, for \( n \ge 4 \), the polynomial \( P(a_1, a_2, \ldots, a_n) \) is irreducible over the real numbers.

The final answer is \( \boxed{ n \ge 4 } \).