Determine for which $ n$ positive integer the equation: $ a \plus{} b \plus{} c \plus{} d \equal{} n \sqrt {abcd}$ has positive integer solutions.
1. **Initial Check for Small Values of \( n \):**
   We start by checking if the equation \( a + b + c + d = n \sqrt{abcd} \) has positive integer solutions for \( n \leq 4 \).

   - For \( n = 1 \):
     \[
     a + b + c + d = \sqrt{abcd}
     \]
     Choosing \( a = b = c = d = 4 \):
     \[
     4 + 4 + 4 + 4 = 16 \quad \text{and} \quad \sqrt{4 \cdot 4 \cdot 4 \cdot 4} = 4
     \]
     \[
     16 = 1 \cdot 4 \quad \text{so} \quad n = 1 \text{ works.}
     \]

   - For \( n = 2 \):
     \[
     a + b + c + d = 2 \sqrt{abcd}
     \]
     Choosing \( a = b = c = d = 2 \):
     \[
     2 + 2 + 2 + 2 = 8 \quad \text{and} \quad \sqrt{2 \cdot 2 \cdot 2 \cdot 2} = 2
     \]
     \[
     8 = 2 \cdot 2 \quad \text{so} \quad n = 2 \text{ works.}
     \]

   - For \( n = 3 \):
     \[
     a + b + c + d = 3 \sqrt{abcd}
     \]
     Choosing \( a = b = 2 \) and \( c = d = 1 \):
     \[
     2 + 2 + 1 + 1 = 6 \quad \text{and} \quad \sqrt{2 \cdot 2 \cdot 1 \cdot 1} = 2
     \]
     \[
     6 = 3 \cdot 2 \quad \text{so} \quad n = 3 \text{ works.}
     \]

   - For \( n = 4 \):
     \[
     a + b + c + d = 4 \sqrt{abcd}
     \]
     Choosing \( a = b = c = d = 1 \):
     \[
     1 + 1 + 1 + 1 = 4 \quad \text{and} \quad \sqrt{1 \cdot 1 \cdot 1 \cdot 1} = 1
     \]
     \[
     4 = 4 \cdot 1 \quad \text{so} \quad n = 4 \text{ works.}
     \]

2. **Claim for \( n \geq 5 \):**
   We claim that there are no positive integer solutions for \( n \geq 5 \).

3. **Squaring Both Sides:**
   Squaring both sides of the equation \( a + b + c + d = n \sqrt{abcd} \), we get:
   \[
   (a + b + c + d)^2 = n^2 abcd
   \]
   Expanding the left-hand side:
   \[
   a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd) = n^2 abcd
   \]

4. **Vieta Jumping:**
   We rewrite the equation as a quadratic in \( a \):
   \[
   a^2 + (2b + 2c + 2d - n^2 bcd)a + (b + c + d)^2 = 0
   \]
   Let \( f(x) = x^2 + (2b + 2c + 2d - n^2 bcd)x + (b + c + d)^2 \). We know \( f(a) = 0 \).

5. **Assumption and Contradiction:**
   Without loss of generality, assume \( a \geq b \geq c \geq d \). Suppose there exists a quadruplet \( (a, b, c, d) \) with minimal sum. Let \( t \) be the other root of \( f(x) = 0 \).

   By Vieta's formulas, the sum of the roots \( a \) and \( t \) is:
   \[
   a + t = -(2b + 2c + 2d - n^2 bcd)
   \]
   Since \( t \) is another root, we have:
   \[
   t = -(2b + 2c + 2d - n^2 bcd) - a
   \]
   Observing that:
   \[
   f(b) = b^2 + (2b + 2c + 2d - n^2 bcd)b + (b + c + d)^2
   \]
   Simplifying:
   \[
   f(b) = b^2 + 2b^2 + 2bc + 2bd - n^2 b^2 cd + (b + c + d)^2
   \]
   \[
   f(b) = (16 - n^2 cd)b^2 < 0 \quad \text{for} \quad n \geq 5
   \]
   This implies \( t < b \leq a \), which contradicts the minimality of the sum of \( (a, b, c, d) \).

6. **Conclusion:**
   Hence, there are no solutions for \( n \geq 5 \).

The final answer is \( \boxed{ n = 1, 2, 3, 4 } \)