Find all positive integers $\overline{xyz}$ ($x$, $y$ and $z$ are digits) such that
$\overline{xyz} = x+y+z+xy+yz+zx+xyz$
1. Start with the given equation:
   \[
   \overline{xyz} = x + y + z + xy + yz + zx + xyz
   \]
   where $\overline{xyz}$ represents the three-digit number $100x + 10y + z$.

2. Substitute $\overline{xyz}$ with $100x + 10y + z$:
   \[
   100x + 10y + z = x + y + z + xy + yz + zx + xyz
   \]

3. Simplify the equation by subtracting $x + y + z$ from both sides:
   \[
   100x + 10y + z - x - y - z = xy + yz + zx + xyz
   \]
   \[
   99x + 9y = xy + yz + zx + xyz
   \]

4. Rearrange the equation to isolate terms involving $x$ and $y$ on one side:
   \[
   99x + 9y = xy + yz + zx + xyz
   \]

5. Factor out common terms:
   \[
   9(11x + y) = xy + yz + zx + xyz
   \]

6. Since $x$, $y$, and $z$ are digits (0 to 9), we can test possible values for $y$ and $z$ to find a solution. Notice that $9 \ge y, z$ implies that $LHS \ge 0$ and $RHS \ge 0$.

7. To satisfy the equation, consider the case where $y = 9$ and $z = 9$:
   \[
   9(11x + 9) = x \cdot 9 + 9 \cdot 9 + 9 \cdot x + x \cdot 9 \cdot 9
   \]
   \[
   9(11x + 9) = 9x + 81 + 9x + 81x
   \]
   \[
   9(11x + 9) = 99x + 81
   \]

8. Simplify the equation:
   \[
   99x + 81 = 99x + 81
   \]

9. This equation holds true for any $x$ from 1 to 9. Therefore, the solutions are:
   \[
   x = 1, 2, 3, \ldots, 9
   \]

10. The corresponding three-digit numbers are:
    \[
    199, 299, 399, \ldots, 999
    \]

The final answer is $\boxed{199, 299, 399, \ldots, 999}$