The positive real numbers $a, b, c$ satisfy the equality $a + b + c = 1$. For every natural number $n$ find the minimal possible value of the expression $$E=\frac{a^{-n}+b}{1-a}+\frac{b^{-n}+c}{1-b}+\frac{c^{-n}+a}{1-c}$$
Given the positive real numbers \(a, b, c\) satisfy the equality \(a + b + c = 1\). We need to find the minimal possible value of the expression 
\[ E = \frac{a^{-n} + b}{1-a} + \frac{b^{-n} + c}{1-b} + \frac{c^{-n} + a}{1-c} \]
for every natural number \(n\).

1. **Rewrite the first term:**
   \[
   \frac{a^{-n} + b}{1-a} = \frac{1 + a^n b}{a^n (b+c)}
   \]
   This is because \(1 - a = b + c\) from the given condition \(a + b + c = 1\).

2. **Simplify the expression:**
   \[
   \frac{1 + a^n b}{a^n (b+c)} = \frac{a^{n+1} + 1 + a^n b - a^{n+1}}{a^n (b+c)} = \frac{a + b}{b+c} + \frac{(b+c)(1 + a + a^2 + \dots + a^n)}{a^n (b+c)}
   \]
   This simplifies to:
   \[
   \frac{a + b}{b+c} + \frac{1 + a + a^2 + \dots + a^n}{a^n}
   \]

3. **Sum the cyclic terms:**
   \[
   E = \sum_{\text{cyc}} \left( \frac{a + b}{b+c} + \frac{1 + a + a^2 + \dots + a^n}{a^n} \right)
   \]

4. **Estimate the first sum using AM-GM inequality:**
   \[
   \sum_{\text{cyc}} \frac{a + b}{b+c} \geq 3
   \]
   This follows from the AM-GM inequality applied to the terms \(\frac{a + b}{b+c}\).

5. **Estimate the second sum using Power Mean inequality:**
   \[
   \left( \frac{a^{-k} + b^{-k} + c^{-k}}{3} \right)^{-\frac{1}{k}} \leq \frac{a + b + c}{3} \implies a^{-k} + b^{-k} + c^{-k} \geq 3^{k+1}
   \]
   Therefore,
   \[
   \sum_{k=1}^{n} \left( \sum_{\text{cyc}} a^{-k} \right) \geq \sum_{k=1}^{n} 3^{k+1}
   \]

6. **Combine the estimates:**
   \[
   E \geq 3 + 3 + 3^2 + \dots + 3^{n+1}
   \]
   The sum of the geometric series \(3 + 3^2 + \dots + 3^{n+1}\) is:
   \[
   3 \left( \frac{3^{n+1} - 1}{3 - 1} \right) = \frac{3^{n+2} - 3}{2}
   \]

7. **Add the initial 3:**
   \[
   E \geq 3 + \frac{3^{n+2} - 3}{2} = \frac{3^{n+2} + 3}{2}
   \]

Thus, the minimal possible value of \(E\) is:
\[
\boxed{\frac{3^{n+2} + 3}{2}}
\]