Let $n$ be a positive integer. Positive numbers $a$, $b$, $c$ satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Find the greatest possible value of $$E(a,b,c)=\frac{a^{n}}{a^{2n+1}+b^{2n} \cdot c + b \cdot c^{2n}}+\frac{b^{n}}{b^{2n+1}+c^{2n} \cdot a + c \cdot a^{2n}}+\frac{c^{n}}{c^{2n+1}+a^{2n} \cdot b + a \cdot b^{2n}}$$
To find the greatest possible value of 
\[ E(a,b,c) = \frac{a^n}{a^{2n+1} + b^{2n}c + bc^{2n}} + \frac{b^n}{b^{2n+1} + c^{2n}a + ca^{2n}} + \frac{c^n}{c^{2n+1} + a^{2n}b + ab^{2n}}, \]
we start by using the given condition:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1. \]

1. **Applying the AM-GM Inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 3 \sqrt[3]{\frac{1}{abc}}. \]
   Given that \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\), it follows that:
   \[ 1 \geq 3 \sqrt[3]{\frac{1}{abc}} \implies \frac{1}{3} \geq \sqrt[3]{\frac{1}{abc}} \implies 27 \geq abc. \]

2. **Using Cauchy-Schwarz Inequality:**
   We apply the Cauchy-Schwarz inequality in the form:
   \[ (a^{2n+1} + b^{2n}c + bc^{2n}) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq (a^n + b^n + c^n)^2. \]
   Given \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\), we have:
   \[ a^{2n+1} + b^{2n}c + bc^{2n} \geq (a^n + b^n + c^n)^2. \]

3. **Analyzing the Expression \(E(a,b,c)\):**
   We need to maximize:
   \[ E(a,b,c) = \frac{a^n}{a^{2n+1} + b^{2n}c + bc^{2n}} + \frac{b^n}{b^{2n+1} + c^{2n}a + ca^{2n}} + \frac{c^n}{c^{2n+1} + a^{2n}b + ab^{2n}}. \]
   Using the inequality derived from Cauchy-Schwarz:
   \[ \frac{a^n}{a^{2n+1} + b^{2n}c + bc^{2n}} \leq \frac{a^n}{(a^n + b^n + c^n)^2}. \]
   Similarly for the other terms, we get:
   \[ E(a,b,c) \leq \frac{a^n}{(a^n + b^n + c^n)^2} + \frac{b^n}{(a^n + b^n + c^n)^2} + \frac{c^n}{(a^n + b^n + c^n)^2}. \]
   Factoring out the common denominator:
   \[ E(a,b,c) \leq \frac{a^n + b^n + c^n}{(a^n + b^n + c^n)^2} = \frac{1}{a^n + b^n + c^n}. \]

4. **Maximizing the Expression:**
   To maximize \(E(a,b,c)\), we need to minimize \(a^n + b^n + c^n\). Given \(a = b = c = 3\) (since \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\) and \(a, b, c\) are positive numbers), we have:
   \[ a^n + b^n + c^n = 3^n + 3^n + 3^n = 3 \cdot 3^n = 3^{n+1}. \]
   Therefore:
   \[ E(a,b,c) \leq \frac{1}{3^{n+1}}. \]

5. **Conclusion:**
   The maximum value of \(E(a,b,c)\) is achieved when \(a = b = c = 3\), giving:
   \[ E(a,b,c) = \frac{1}{3^{n+1}}. \]

The final answer is \(\boxed{\frac{1}{3^{n+1}}}\).