Find all positive integers n such that  $ (n ^{ 2} + 11n - 4) n! + 33.13 ^ n + 4 $ is the perfect square 
We need to find all positive integers \( n \) such that \( \left(n^2 + 11n - 4\right)n! + 33 \times 13^n + 4 \) is a perfect square. Let \( m = \left(n^2 + 11n - 4\right)n! + 33 \times 13^n + 4 \).

We will consider two cases based on the value of \( n \).

### Case 1: \( n \geq 7 \)

First, we analyze the expression modulo 7. The quadratic residues modulo 7 are \( \{0, 1, 2, 4\} \). Since \( n! \equiv 0 \pmod{7} \) for \( n \geq 7 \), we can write:
\[
m = \left(n^2 + 11n - 4\right)n! + 33 \times 13^n + 4 \equiv -2 \times (-1)^n + 4 \pmod{7}
\]
This simplifies to:
\[
m \equiv 2 \pmod{7} \quad \text{if } n \text{ is even}
\]
\[
m \equiv 6 \pmod{7} \quad \text{if } n \text{ is odd}
\]
Since 6 is not a quadratic residue modulo 7, \( n \) must be even. Let \( n = 2k \) for some positive integer \( k \).

Next, we analyze the expression modulo 8. Since \( n \geq 7 \), \( n! \equiv 0 \pmod{8} \) and the quadratic residues modulo 8 are \( \{0, 1, 4\} \). We have:
\[
m = \left(n^2 + 11n - 4\right)n! + 33 \times 13^n + 4 \equiv 5^n + 4 \pmod{8}
\]
Since \( n \) is even, let \( n = 2k \):
\[
m \equiv (5^2)^k + 4 \equiv 25^k + 4 \equiv 1^k + 4 \equiv 5 \pmod{8}
\]
But 5 is not a quadratic residue modulo 8. Thus, if \( n \geq 7 \), \( m \) cannot be a perfect square. This is a contradiction, so \( n < 7 \).

### Case 2: \( n < 7 \)

We now check each value of \( n \) from 1 to 6 to see if \( m \) is a perfect square.

- For \( n = 1 \):
\[
m = (1^2 + 11 \cdot 1 - 4) \cdot 1! + 33 \cdot 13^1 + 4 = (1 + 11 - 4) \cdot 1 + 33 \cdot 13 + 4 = 8 + 429 + 4 = 441 = 21^2
\]
So, \( m \) is a perfect square.

- For \( n = 2 \):
\[
m = (2^2 + 11 \cdot 2 - 4) \cdot 2! + 33 \cdot 13^2 + 4 = (4 + 22 - 4) \cdot 2 + 33 \cdot 169 + 4 = 22 \cdot 2 + 5577 + 4 = 44 + 5577 + 4 = 5625 = 75^2
\]
So, \( m \) is a perfect square.

- For \( n = 3 \):
\[
m = (3^2 + 11 \cdot 3 - 4) \cdot 3! + 33 \cdot 13^3 + 4 = (9 + 33 - 4) \cdot 6 + 33 \cdot 2197 + 4 = 38 \cdot 6 + 72501 + 4 = 228 + 72501 + 4 = 72733
\]
\( 72733 \) is not a perfect square.

- For \( n = 4 \):
\[
m = (4^2 + 11 \cdot 4 - 4) \cdot 4! + 33 \cdot 13^4 + 4 = (16 + 44 - 4) \cdot 24 + 33 \cdot 28561 + 4 = 56 \cdot 24 + 942513 + 4 = 1344 + 942513 + 4 = 943861
\]
\( 943861 \) is not a perfect square.

- For \( n = 5 \):
\[
m = (5^2 + 11 \cdot 5 - 4) \cdot 5! + 33 \cdot 13^5 + 4 = (25 + 55 - 4) \cdot 120 + 33 \cdot 371293 + 4 = 76 \cdot 120 + 12252669 + 4 = 9120 + 12252669 + 4 = 12261793
\]
\( 12261793 \) is not a perfect square.

- For \( n = 6 \):
\[
m = (6^2 + 11 \cdot 6 - 4) \cdot 6! + 33 \cdot 13^6 + 4 = (36 + 66 - 4) \cdot 720 + 33 \cdot 4826809 + 4 = 98 \cdot 720 + 159284697 + 4 = 70560 + 159284697 + 4 = 159355261
\]
\( 159355261 \) is not a perfect square.

Thus, the only positive integers \( n \) that make \( m \) a perfect square are \( n = 1 \) and \( n = 2 \).

The final answer is \( \boxed{(n, m) = \{(1, 21^2), (2, 75^2)\}} \).