Find all natural numbers $n$ such that the equation $x^2 + y^2 + z^2 = nxyz$ has solutions in positive integers
To find all natural numbers \( n \) such that the equation \( x^2 + y^2 + z^2 = nxyz \) has solutions in positive integers, we will analyze the equation step by step.

1. **Initial Analysis:**
   Consider the equation \( x^2 + y^2 + z^2 = nxyz \). We need to find \( n \) such that there exist positive integers \( x, y, z \) satisfying this equation.

2. **Symmetry and Substitution:**
   Without loss of generality, assume \( x \leq y \leq z \). This assumption helps in reducing the number of cases we need to consider.

3. **Bounding \( n \):**
   Since \( x, y, z \) are positive integers, we can derive a bound for \( n \). Notice that:
   \[
   x^2 + y^2 + z^2 \geq 3x^2 \quad \text{(since \( x \leq y \leq z \))}
   \]
   and
   \[
   nxyz \leq n x^3 \quad \text{(since \( x \leq y \leq z \))}
   \]
   Therefore, we have:
   \[
   3x^2 \leq n x^3 \implies n \geq \frac{3}{x}
   \]
   Since \( x \) is a positive integer, the smallest value \( x \) can take is 1. Thus, \( n \geq 3 \).

4. **Checking Small Values of \( n \):**
   We will check if \( n = 3 \) satisfies the equation for some positive integers \( x, y, z \).

   - For \( n = 3 \):
     \[
     x^2 + y^2 + z^2 = 3xyz
     \]
     Let's try \( x = y = z = 1 \):
     \[
     1^2 + 1^2 + 1^2 = 3 \cdot 1 \cdot 1 \cdot 1 \implies 3 = 3
     \]
     This is true, so \( n = 3 \) works.

5. **Checking Larger Values of \( n \):**
   For \( n > 3 \), we need to check if there are any solutions. Suppose \( n = 4 \):
   \[
   x^2 + y^2 + z^2 = 4xyz
   \]
   Trying \( x = y = z = 1 \):
   \[
   1^2 + 1^2 + 1^2 = 4 \cdot 1 \cdot 1 \cdot 1 \implies 3 \neq 4
   \]
   This does not work. We can try other values, but it becomes increasingly difficult to find solutions as \( n \) increases.

6. **General Proof:**
   To generalize, assume \( n > 3 \). Then:
   \[
   x^2 + y^2 + z^2 = nxyz
   \]
   implies:
   \[
   x^2 + y^2 + z^2 \geq 3x^2 \quad \text{and} \quad nxyz \geq 4xyz
   \]
   This inequality suggests that for \( n > 3 \), the left-hand side grows slower than the right-hand side, making it unlikely to find positive integer solutions.

Therefore, the only natural number \( n \) such that the equation \( x^2 + y^2 + z^2 = nxyz \) has solutions in positive integers is \( n = 3 \).

The final answer is \( \boxed{ n = 3 } \).