Determine the minimum value of prime $p> 3$ for which there is no natural number $n> 0$ such that $2^n+3^n\equiv 0\pmod{p} $.
1. **Identify the conditions for \( p \) to satisfy the given congruence:**
   We need to find the minimum prime \( p > 3 \) such that there is no natural number \( n > 0 \) for which \( 2^n + 3^n \equiv 0 \pmod{p} \).

2. **Analyze the quadratic residues:**
   For \( 2^n + 3^n \equiv 0 \pmod{p} \), both \( 2 \) and \( 3 \) must either be quadratic residues or non-quadratic residues modulo \( p \). According to the law of quadratic reciprocity, this occurs if and only if \( p \equiv \pm 1, \pm 5 \pmod{24} \).

3. **Check the smallest primes \( p \equiv \pm 1, \pm 5 \pmod{24} \):**
   - The smallest prime \( p > 3 \) that fits these conditions is \( p = 5 \).
   - Verify if \( p = 5 \) satisfies the given conditions:
     \[
     2^n + 3^n \equiv 0 \pmod{5}
     \]
     - For \( n = 1 \):
       \[
       2^1 + 3^1 = 2 + 3 = 5 \equiv 0 \pmod{5}
       \]
     - Since \( n = 1 \) is a solution, \( p = 5 \) does not satisfy the given conditions.

4. **Check the next smallest prime \( p \equiv \pm 1, \pm 5 \pmod{24} \):**
   - The next smallest prime is \( p = 19 \).
   - Verify if \( p = 19 \) satisfies the given conditions:
     \[
     2^n + 3^n \equiv 0 \pmod{19}
     \]
     - Rewrite the equation:
       \[
       2^n \equiv -3^n \pmod{19}
       \]
     - Multiply both sides by \( 3^{-n} \) (the modular inverse of \( 3^n \) modulo \( 19 \)):
       \[
       \left(2 \cdot 3^{-1}\right)^n \equiv -1 \pmod{19}
       \]
     - Calculate \( 3^{-1} \mod 19 \):
       \[
       3 \cdot 13 \equiv 1 \pmod{19} \implies 3^{-1} \equiv 13 \pmod{19}
       \]
     - Substitute \( 3^{-1} \):
       \[
       (2 \cdot 13)^n \equiv -1 \pmod{19} \implies 26^n \equiv -1 \pmod{19}
       \]
     - Simplify \( 26 \mod 19 \):
       \[
       26 \equiv 7 \pmod{19}
       \]
     - Thus, the equation becomes:
       \[
       7^n \equiv -1 \pmod{19}
       \]
     - Check if there exists any \( n \) such that \( 7^n \equiv -1 \pmod{19} \):
       - \( 7^1 \equiv 7 \pmod{19} \)
       - \( 7^2 \equiv 49 \equiv 11 \pmod{19} \)
       - \( 7^3 \equiv 7 \cdot 11 \equiv 77 \equiv 1 \pmod{19} \)
       - \( 7^4 \equiv 7 \cdot 1 \equiv 7 \pmod{19} \)
       - Continue this process, and it is evident that \( 7^n \) cycles through values but never equals \( -1 \pmod{19} \).

5. **Conclusion:**
   Since there is no \( n \) such that \( 7^n \equiv -1 \pmod{19} \), \( p = 19 \) satisfies the given conditions.

The final answer is \( \boxed{19} \)