Let $ a,b,c$ be non-zero real numbers satisfying \[ \dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\equal{}\dfrac{1}{a\plus{}b\plus{}c}.\] Find all integers $ n$ such that \[ \dfrac{1}{a^n}\plus{}\dfrac{1}{b^n}\plus{}\dfrac{1}{c^n}\equal{}\dfrac{1}{a^n\plus{}b^n\plus{}c^n}.\]
1. Given the equation:
   \[
   \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}
   \]
   we start by multiplying both sides by \(abc(a+b+c)\) to clear the denominators:
   \[
   abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = abc \cdot \frac{1}{a+b+c}
   \]
   Simplifying the left-hand side:
   \[
   bc(a+b+c) + ac(a+b+c) + ab(a+b+c) = abc
   \]
   This simplifies to:
   \[
   (a+b+c)(ab+bc+ca) = abc
   \]
2. Expanding the left-hand side:
   \[
   a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc = abc
   \]
   Subtracting \(abc\) from both sides:
   \[
   a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 2abc = 0
   \]
3. Factoring the left-hand side:
   \[
   (a+b)(b+c)(c+a) = 0
   \]
   This implies that at least one of the factors must be zero. Without loss of generality, assume:
   \[
   a + b = 0
   \]
   Then, \(a = -b\). Substituting \(a = -b\) into the original equation:
   \[
   \frac{1}{-b} + \frac{1}{b} + \frac{1}{c} = \frac{1}{-b + b + c}
   \]
   Simplifying:
   \[
   -\frac{1}{b} + \frac{1}{b} + \frac{1}{c} = \frac{1}{c}
   \]
   This is always true, so the condition \(a + b = 0\) is valid.

4. Now, consider the equation for \(n\):
   \[
   \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{a^n + b^n + c^n}
   \]
   Substituting \(a = -b\):
   \[
   \frac{1}{(-b)^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{(-b)^n + b^n + c^n}
   \]
   Simplifying:
   \[
   \frac{(-1)^n}{b^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{(-1)^n b^n + b^n + c^n}
   \]
   If \(n\) is odd, \((-1)^n = -1\), so:
   \[
   -\frac{1}{b^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{-b^n + b^n + c^n}
   \]
   This simplifies to:
   \[
   \frac{1}{c^n} = \frac{1}{c^n}
   \]
   which is always true.

5. If \(n\) is even, \((-1)^n = 1\), so:
   \[
   \frac{1}{b^n} + \frac{1}{b^n} + \frac{1}{c^n} = \frac{1}{b^n + b^n + c^n}
   \]
   Simplifying:
   \[
   \frac{2}{b^n} + \frac{1}{c^n} = \frac{1}{2b^n + c^n}
   \]
   This is not generally true for non-zero \(b\) and \(c\).

Therefore, the solution is valid only for odd \(n\).

The final answer is \( \boxed{ n = 2k - 1 } \) where \( k \in \mathbb{N} \).