For every positive integer $n$, determine the greatest possible value of the quotient
$$\frac{1-x^{n}-(1-x)^{n}}{x(1-x)^n+(1-x)x^n}$$
where $0 < x < 1$.
1. We start with the given expression:
   \[
   \frac{1 - x^n - (1 - x)^n}{x(1 - x)^n + (1 - x)x^n}
   \]
   where \(0 < x < 1\).

2. Let \(x = \frac{a}{a+b}\) and \(y = \frac{b}{a+b}\). Note that \(x + y = 1\). We need to transform the given expression using these substitutions.

3. Substitute \(x = \frac{a}{a+b}\) and \(y = \frac{b}{a+b}\) into the expression:
   \[
   \frac{1 - \left(\frac{a}{a+b}\right)^n - \left(\frac{b}{a+b}\right)^n}{\left(\frac{a}{a+b}\right)\left(\frac{b}{a+b}\right)^n + \left(\frac{b}{a+b}\right)\left(\frac{a}{a+b}\right)^n}
   \]

4. Simplify the numerator:
   \[
   1 - \left(\frac{a}{a+b}\right)^n - \left(\frac{b}{a+b}\right)^n = \frac{(a+b)^n - a^n - b^n}{(a+b)^n}
   \]

5. Simplify the denominator:
   \[
   \left(\frac{a}{a+b}\right)\left(\frac{b}{a+b}\right)^n + \left(\frac{b}{a+b}\right)\left(\frac{a}{a+b}\right)^n = \frac{ab^n + a^n b}{(a+b)^{n+1}}
   \]

6. Combine the simplified numerator and denominator:
   \[
   \frac{\frac{(a+b)^n - a^n - b^n}{(a+b)^n}}{\frac{ab^n + a^n b}{(a+b)^{n+1}}} = \frac{(a+b)^{n+1} - a^n(a+b) - b^n(a+b)}{ab^n + a^n b}
   \]

7. We need to show that:
   \[
   \frac{(a+b)^{n+1} - a^n(a+b) - b^n(a+b)}{ab^n + a^n b} \leq 2^n - 2
   \]

8. Simplify the inequality:
   \[
   (a+b)^{n+1} - a^n(a+b) - b^n(a+b) \leq (2^n - 2)(ab^n + a^n b)
   \]

9. Expand and rearrange terms:
   \[
   \sum_{i=1}^{n} a^i b^{n+1-i} \binom{n+1}{i} - a^n b - ab^n \leq (2^n - 2)(ab^n + a^n b)
   \]

10. Simplify further:
    \[
    \frac{1}{2} \sum_{i=1}^{n} \left( a^i b^{n+1-i} + a^{n+1-i} b^i \right) \binom{n+1}{i} \leq (2^n - 1)(ab^n + a^n b)
    \]

11. Since:
    \[
    \frac{1}{2} \sum_{i=1}^{n} \binom{n+1}{i} \leq \frac{1}{2}(2^{n+1} - 2) = 2^n - 1
    \]

12. It suffices to prove that:
    \[
    a^i b^{n+1-i} + a^{n+1-i} b^i \geq ab^n + a^n b \quad \text{for all} \quad i = 1, 2, \ldots, n
    \]

13. This is equivalent to:
    \[
    (a^{n+1-i} - b^{n+1-i})(a^i - b^i) = (a - b)^2 \left( \sum_{k=0}^{n-i} a^k b^{n-i-k} \right) \left( \sum_{j=0}^{i-1} a^j b^{i-1-j} \right) \geq 0
    \]

14. Since \((a - b)^2 \geq 0\) and the sums are non-negative, the inequality holds.

\(\blacksquare\)

The final answer is \( \boxed{ 2^n - 2 } \)