Find the smallest positive integer $n,$ such that   $3^k+n^k+ (3n)^k+ 2014^k$ is a perfect square for all natural numbers $k,$ but not a perfect cube, for all natural numbers $k.$
1. **Verify \( n = 1 \):**
   \[
   3^k + n^k + (3n)^k + 2014^k = 3^k + 1^k + 3^k + 2014^k = 2 \cdot 3^k + 1 + 2014^k
   \]
   Consider the expression modulo 3:
   \[
   2 \cdot 3^k + 1 + 2014^k \equiv 0 + 1 + 1 \equiv 2 \pmod{3}
   \]
   Since 2 modulo 3 is not a quadratic residue, \( 2 \cdot 3^k + 1 + 2014^k \) is never a perfect square. Therefore, \( n = 1 \) does not work.

2. **Verify \( n = 2 \):**
   \[
   3^k + n^k + (3n)^k + 2014^k = 3^k + 2^k + 6^k + 2014^k
   \]
   For \( k = 1 \):
   \[
   3^1 + 2^1 + 6^1 + 2014^1 = 3 + 2 + 6 + 2014 = 2025 = 45^2
   \]
   This shows that for \( k = 1 \), the expression is a perfect square.

3. **Check modulo 7 for all \( k \):**
   \[
   3^k + 2^k + 6^k + 2014^k \pmod{7}
   \]
   Since \( 2014 \equiv 2 \pmod{7} \), we have:
   \[
   2014^k \equiv 2^k \pmod{7}
   \]
   Therefore:
   \[
   3^k + 2^k + 6^k + 2^k \equiv 3^k + 2 \cdot 2^k + 6^k \pmod{7}
   \]
   We need to check the residues of \( 3^k, 2^k, \) and \( 6^k \) modulo 7:
   \[
   3^k \equiv 3, 2, 6, 4, 5, 1 \pmod{7} \quad \text{(repeats every 6 terms)}
   \]
   \[
   2^k \equiv 2, 4, 1 \pmod{7} \quad \text{(repeats every 3 terms)}
   \]
   \[
   6^k \equiv 6, 1 \pmod{7} \quad \text{(repeats every 2 terms)}
   \]
   Combining these, we get:
   \[
   3^k + 2 \cdot 2^k + 6^k \equiv 3, 4, 5, 2, 4, 4, 2, 4, 5, 2, 4, 4, \ldots \pmod{7}
   \]
   A perfect cube modulo 7 can only be \( 0, 1, \) or \( 6 \). The sequence above does not match these values, so \( 3^k + 2^k + 6^k + 2014^k \) is never a perfect cube.

Therefore, \( n = 2 \) works and is the smallest positive integer satisfying the conditions.

The final answer is \( \boxed{2} \).