Find all positive integers $n$ such that $(n^2 + 11n - 4) \cdot  n! + 33 \cdot 13^n + 4$ is a perfect square
To find all positive integers \( n \) such that \((n^2 + 11n - 4) \cdot n! + 33 \cdot 13^n + 4\) is a perfect square, we will use modular arithmetic and check small values of \( n \).

1. **Check for small values of \( n \):**
   - For \( n = 1 \):
     \[
     (1^2 + 11 \cdot 1 - 4) \cdot 1! + 33 \cdot 13^1 + 4 = (1 + 11 - 4) \cdot 1 + 33 \cdot 13 + 4 = 8 + 429 + 4 = 441
     \]
     \( 441 = 21^2 \), which is a perfect square.
   - For \( n = 2 \):
     \[
     (2^2 + 11 \cdot 2 - 4) \cdot 2! + 33 \cdot 13^2 + 4 = (4 + 22 - 4) \cdot 2 + 33 \cdot 169 + 4 = 22 \cdot 2 + 5577 + 4 = 44 + 5577 + 4 = 5625
     \]
     \( 5625 = 75^2 \), which is a perfect square.
   - For \( n = 3 \):
     \[
     (3^2 + 11 \cdot 3 - 4) \cdot 3! + 33 \cdot 13^3 + 4 = (9 + 33 - 4) \cdot 6 + 33 \cdot 2197 + 4 = 38 \cdot 6 + 72501 + 4 = 228 + 72501 + 4 = 72733
     \]
     \( 72733 \) is not a perfect square.
   - For \( n = 4 \):
     \[
     (4^2 + 11 \cdot 4 - 4) \cdot 4! + 33 \cdot 13^4 + 4 = (16 + 44 - 4) \cdot 24 + 33 \cdot 28561 + 4 = 56 \cdot 24 + 942513 + 4 = 1344 + 942513 + 4 = 943861
     \]
     \( 943861 \) is not a perfect square.
   - For \( n = 5 \):
     \[
     (5^2 + 11 \cdot 5 - 4) \cdot 5! + 33 \cdot 13^5 + 4 = (25 + 55 - 4) \cdot 120 + 33 \cdot 371293 + 4 = 76 \cdot 120 + 12252669 + 4 = 9120 + 12252669 + 4 = 12261793
     \]
     \( 12261793 \) is not a perfect square.
   - For \( n = 6 \):
     \[
     (6^2 + 11 \cdot 6 - 4) \cdot 6! + 33 \cdot 13^6 + 4 = (36 + 66 - 4) \cdot 720 + 33 \cdot 4826809 + 4 = 98 \cdot 720 + 159284697 + 4 = 70560 + 159284697 + 4 = 159355261
     \]
     \( 159355261 \) is not a perfect square.
   - For \( n = 7 \):
     \[
     (7^2 + 11 \cdot 7 - 4) \cdot 7! + 33 \cdot 13^7 + 4 = (49 + 77 - 4) \cdot 5040 + 33 \cdot 62748517 + 4 = 122 \cdot 5040 + 2070701061 + 4 = 614880 + 2070701061 + 4 = 2071315945
     \]
     \( 2071315945 \) is not a perfect square.

2. **Use modular arithmetic for \( n \geq 8 \):**
   - Consider the expression modulo 7:
     \[
     (n^2 + 11n - 4) \cdot n! + 33 \cdot 13^n + 4 \pmod{7}
     \]
     Since \( 13 \equiv -1 \pmod{7} \), we have:
     \[
     33 \cdot 13^n \equiv 33 \cdot (-1)^n \equiv 5 \cdot (-1)^n \pmod{7}
     \]
     For \( n \geq 7 \), \( n! \equiv 0 \pmod{7} \), so:
     \[
     (n^2 + 11n - 4) \cdot n! \equiv 0 \pmod{7}
     \]
     Thus:
     \[
     5 \cdot (-1)^n + 4 \equiv k^2 \pmod{7}
     \]
     For \( n \) even:
     \[
     5 \cdot 1 + 4 \equiv 9 \equiv 2 \pmod{7}
     \]
     For \( n \) odd:
     \[
     5 \cdot (-1) + 4 \equiv -5 + 4 \equiv -1 \equiv 6 \pmod{7}
     \]
     Neither 2 nor 6 are quadratic residues modulo 7, so there are no solutions for \( n \geq 8 \).

Conclusion:
The only solutions are \( n = 1 \) and \( n = 2 \).

The final answer is \( \boxed{ n = \{1, 2\} } \).