Find all positive integer $n$ such that the equation $x^3+y^3+z^3=n \cdot x^2 \cdot y^2 \cdot z^2$ has positive integer solutions.
To find all positive integers \( n \) such that the equation \( x^3 + y^3 + z^3 = n \cdot x^2 \cdot y^2 \cdot z^2 \) has positive integer solutions, we will proceed with the following steps:

1. **Assume \( x \geq y \geq z \):**
   \[
   x^3 + y^3 + z^3 \leq 3x^3
   \]
   Therefore,
   \[
   3x^3 \geq n \cdot x^2 \cdot y^2 \cdot z^2
   \]
   Simplifying, we get:
   \[
   x \geq \frac{n \cdot y^2 \cdot z^2}{3}
   \]

2. **Divisibility Condition:**
   Since \( x^3 + y^3 + z^3 = n \cdot x^2 \cdot y^2 \cdot z^2 \), we have:
   \[
   x^2 \mid y^3 + z^3
   \]
   Given \( y \geq z \), we have:
   \[
   2y^3 \geq y^3 + z^3 \geq x^2 \geq \frac{n^2 \cdot y^4 \cdot z^4}{9}
   \]

3. **Case Analysis for \( z \):**
   - **If \( z > 1 \):**
     \[
     y \leq \frac{18}{n^2}
     \]
     Since \( y \) is a positive integer, the only possible value for \( y \) is 1. However, this contradicts \( y \geq z \) because \( z > 1 \).

   - **If \( z = 1 \):**
     \[
     y^3 + 1 \geq \frac{n^2 \cdot y^4}{9}
     \]
     Simplifying, we get:
     \[
     9 + \frac{9}{y^3} \geq n^2 y
     \]

4. **Subcase Analysis for \( y \):**
   - **If \( y = 1 \):**
     \[
     x^3 + 2 = n \cdot x^2
     \]
     Solving for \( x \):
     \[
     x^2 \mid 2 \implies x = 1
     \]
     Substituting \( x = 1 \):
     \[
     1^3 + 1^3 + 1^3 = n \cdot 1^2 \cdot 1^2 \cdot 1^2 \implies 3 = n
     \]
     Thus, \( n = 3 \) satisfies the equation with the solution \((x, y, z, n) = (1, 1, 1, 3)\).

   - **If \( y > 1 \):**
     \[
     10 \geq n^2 y
     \]
     - **For \( n = 1 \):**
       \[
       y = 2 \implies (x, y, z, n) = (3, 2, 1, 1)
       \]
       Substituting:
       \[
       3^3 + 2^3 + 1^3 = 1 \cdot 3^2 \cdot 2^2 \cdot 1^2 \implies 27 + 8 + 1 = 36
       \]
       This is true, so \( n = 1 \) satisfies the equation.

     - **For \( n = 2 \):**
       \[
       y = 2 \text{ or } y = 1
       \]
       Checking both cases, there are no solutions.

     - **For \( n = 3 \):**
       \[
       y = 1
       \]
       This gives the solution \((1, 1, 1, 3)\), which we already found.

Conclusion:
The only values of \( n \) that satisfy the equation are \( n = 1 \) and \( n = 3 \).

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