Which integers can be represented as \[\frac{(x+y+z)^{2}}{xyz}\] where $x$, $y$, and $z$ are positive integers?
1. **Define the problem and set up the equation:**
   We need to determine which integers \( k \) can be represented as:
   \[
   k = \frac{(x+y+z)^2}{xyz}
   \]
   where \( x, y, z \) are positive integers. Let \( S_k \) be the set of all triples of positive integers \( (a, b, c) \) such that \( k = \frac{(a+b+c)^2}{abc} \).

2. **Minimize the sum \( a + b + c \):**
   Suppose \( S_k \) is non-empty. Take any \( (a, b, c) \in S_k \) such that \( a + b + c \) is minimized. Without loss of generality, assume \( a \leq b \leq c \).

3. **Formulate the quadratic equation:**
   Note that:
   \[
   a^2 - a(kbc - 2b - 2c) + (b + c)^2 = 0
   \]
   This quadratic equation in \( a \) has two solutions, one of which is \( a \). Let the other solution be \( r \).

4. **Apply Vieta's formulas:**
   By Vieta's formulas, we have:
   \[
   ar = (b + c)^2 \quad \text{and} \quad a + r = kbc - 2b - 2c
   \]
   Since \( r \geq a \) (otherwise \( (r, b, c) \in S_k \) and \( r + b + c < a + b + c \), contradicting the minimality of \( a + b + c \)), we have:
   \[
   a \leq b \leq \frac{b + c}{2} \leq c
   \]

5. **Derive inequalities:**
   From \( ar = (b + c)^2 \), we get:
   \[
   a^2 \leq ar = (b + c)^2 \implies a \leq b + c
   \]
   Also, since \( r \leq 2(b + c) \), we have:
   \[
   kbc - 2b - 2c = a + r \leq (b + c) + 2(b + c) = 3(b + c)
   \]
   Rearranging, we get:
   \[
   (kb - 5)(kc - 5) \leq 25
   \]
   This implies that at least one of \( |kb - 5| \) or \( |kc - 5| \leq 5 \), giving \( 0 \leq kb \leq 10 \). Since \( k \neq 0 \) and \( bk \geq k \), we have:
   \[
   1 \leq k \leq 10
   \]

6. **Check specific values of \( k \):**
   - For \( k = 10 \):
     \[
     10b - 5 \quad \text{and} \quad 10c - 5 \quad \text{are both nonzero}
     \]
     Since \( 10b - 5 \geq 5 \) and \( 10c - 5 \geq 5 \), we get:
     \[
     (10b - 5)(10c - 5) \geq 25
     \]
     This implies \( b = c = 1 \), but then:
     \[
     10 = \frac{(a + 1 + 1)^2}{a} = a + 2 + \frac{1}{a}
     \]
     Hence, \( a = 1 \), which is a contradiction. Thus, \( k \neq 10 \).

   - For \( k = 7 \):
     \[
     7b - 5 \quad \text{and} \quad 7c - 5 \quad \text{are both nonzero}
     \]
     If \( c \geq 2 \), then:
     \[
     7b - 5 \geq 7c - 5 \geq 9
     \]
     This leads to:
     \[
     25 \geq (7b - 5)(7c - 5) \geq 81
     \]
     which is a contradiction. Hence, \( c = 1 \), so:
     \[
     7b - 5 \leq \frac{25}{7 \cdot 1 - 5} < 13
     \]
     Therefore, \( b < \frac{18}{7} \leq 3 \), so \( b = 1 \) or \( b = 2 \). If \( b = 1 \):
     \[
     7 = \frac{(a + 1 + 1)^2}{a} = a + 2 + \frac{1}{a}
     \]
     implying \( a = 1 \), which is a contradiction. If \( b = 2 \):
     \[
     7 = \frac{(a + 3)^2}{2a}
     \]
     leading to:
     \[
     a^2 + 6a + 9 = 14a \implies a^2 - 8a + 9 = 0
     \]
     The discriminant of this quadratic is \( 64 - 4 \cdot 9 = 28 \), which is not a perfect square, so \( a \) is not an integer. Hence, \( k \neq 7 \).

7. **Verify solutions for \( k \) between 1 and 9:**
   - \( k = 1 \): \( (9, 9, 9) \)
   - \( k = 2 \): \( (4, 4, 8) \)
   - \( k = 3 \): \( (3, 3, 3) \)
   - \( k = 4 \): \( (2, 2, 4) \)
   - \( k = 5 \): \( (1, 4, 5) \)
   - \( k = 6 \): \( (1, 2, 3) \)
   - \( k = 8 \): \( (1, 1, 2) \)
   - \( k = 9 \): \( (1, 1, 1) \)

Hence, the set of all integers that can be taken on by \( \frac{(a + b + c)^2}{abc} \) is \( \{1, 2, 3, 4, 5, 6, 8, 9\} \).

The final answer is \( \boxed{ \{1, 2, 3, 4, 5, 6, 8, 9\} } \)