Consider (3-variable) polynomials
\[P_n(x,y,z)=(x-y)^{2n}(y-z)^{2n}+(y-z)^{2n}(z-x)^{2n}+(z-x)^{2n}(x-y)^{2n}\]
and
\[Q_n(x,y,z)=[(x-y)^{2n}+(y-z)^{2n}+(z-x)^{2n}]^{2n}.\]
Determine all positive integers $n$ such that the quotient $Q_n(x,y,z)/P_n(x,y,z)$ is a (3-variable) polynomial with rational coefficients.
1. **Define the polynomials \( P_n(x,y,z) \) and \( Q_n(x,y,z) \):**
   \[
   P_n(x,y,z) = (x-y)^{2n}(y-z)^{2n} + (y-z)^{2n}(z-x)^{2n} + (z-x)^{2n}(x-y)^{2n}
   \]
   \[
   Q_n(x,y,z) = [(x-y)^{2n} + (y-z)^{2n} + (z-x)^{2n}]^{2n}
   \]

2. **Determine when the quotient \( \frac{Q_n(x,y,z)}{P_n(x,y,z)} \) is a polynomial with rational coefficients:**
   By Gauss's Lemma, if \( \frac{Q_n(x, y, z)}{P_n(x, y, z)} \in \mathbb{Q}[x, y, z] \), then \( \frac{Q_n(x, y, z)}{P_n(x, y, z)} \in \mathbb{Z}[x, y, z] \).

3. **Evaluate the polynomials at specific values:**
   Let \( (x, y, z) = (1, 2, 3) \):
   \[
   P_n(1, 2, 3) = (1-2)^{2n}(2-3)^{2n} + (2-3)^{2n}(3-1)^{2n} + (3-1)^{2n}(1-2)^{2n}
   \]
   \[
   = (-1)^{2n}(-1)^{2n} + (-1)^{2n}(2)^{2n} + (2)^{2n}(-1)^{2n}
   \]
   \[
   = 1 + 2^{2n} + 2^{2n} = 1 + 2 \cdot 2^{2n} = 1 + 2^{2n+1}
   \]

   \[
   Q_n(1, 2, 3) = [(1-2)^{2n} + (2-3)^{2n} + (3-1)^{2n}]^{2n}
   \]
   \[
   = [(-1)^{2n} + (-1)^{2n} + (2)^{2n}]^{2n}
   \]
   \[
   = [1 + 1 + 2^{2n}]^{2n} = (2 + 2^{2n})^{2n}
   \]

4. **Simplify the quotient:**
   \[
   \frac{Q_n(1, 2, 3)}{P_n(1, 2, 3)} = \frac{(2 + 2^{2n})^{2n}}{1 + 2^{2n+1}}
   \]

5. **Analyze the divisibility condition:**
   We need \( 1 + 2^{2n+1} \) to divide \( (2 + 2^{2n})^{2n} \):
   \[
   1 + 2^{2n+1} \mid (2 + 2^{2n})^{2n}
   \]

6. **Use properties of exponents and divisibility:**
   Consider the prime factorization and properties of powers:
   \[
   1 + 2^{2n+1} \mid 3^{2n}
   \]
   By Zsigmondy's theorem, for \( n > 1 \), \( 2^{2n+1} + 1 \) has a prime factor that does not divide any smaller power of 2, implying it cannot be a power of 3.

7. **Use the Lifting The Exponent (LTE) lemma:**
   \[
   v_3(2^{2n+1} + 1) = 1 + v_3(2n + 1)
   \]
   We need to show:
   \[
   3^{1 + v_3(2n + 1)} < 2^{2n+1} + 1
   \]
   For \( n > 1 \):
   \[
   3^{1 + v_3(2n + 1)} \le 6n + 3 < 2^{2n+1} + 1
   \]

8. **Conclude the only valid \( n \):**
   For \( n = 1 \):
   \[
   P_1(x, y, z) = (x-y)^2(y-z)^2 + (y-z)^2(z-x)^2 + (z-x)^2(x-y)^2
   \]
   \[
   Q_1(x, y, z) = [(x-y)^2 + (y-z)^2 + (z-x)^2]^2
   \]
   \[
   \frac{Q_1(x, y, z)}{P_1(x, y, z)} = 1
   \]
   Therefore, the only \( n \) that works is \( \boxed{n = 1} \).