Let $x$, $y$, $z$ be positive integers satisfying $x<y<z$ and $x+xy+xyz=37$. Find the greatest possible value of $x+y+z$.
1. Given the equation \( x + xy + xyz = 37 \), we start by factoring out \( x \):
   \[
   x(1 + y + yz) = 37
   \]
   Since \( x \), \( y \), and \( z \) are positive integers and \( x < y < z \), \( x \) must be a divisor of 37. The divisors of 37 are 1 and 37, but since \( x < y < z \), \( x \) must be 1 because 37 is too large to satisfy \( x < y < z \).

2. Substituting \( x = 1 \) into the equation:
   \[
   1(1 + y + yz) = 37 \implies 1 + y + yz = 37 \implies y + yz = 36 \implies y(z + 1) = 36
   \]

3. We need to find pairs \((y, z)\) such that \( y(z + 1) = 36 \) and \( y < z \). We list the factor pairs of 36:
   \[
   (y, z+1) = (1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (9, 4), (12, 3), (18, 2), (36, 1)
   \]
   We discard pairs where \( y \geq z+1 \) or where \( y \) and \( z \) do not satisfy \( y < z \).

4. Valid pairs \((y, z+1)\) where \( y < z \):
   \[
   (2, 18), (3, 12), (4, 9)
   \]
   From these pairs, we derive \( z \) by subtracting 1 from the second element:
   \[
   (y, z) = (2, 17), (3, 11), (4, 8)
   \]

5. Calculate \( x + y + z \) for each valid pair:
   \[
   \begin{align*}
   (x, y, z) = (1, 2, 17) & \implies x + y + z = 1 + 2 + 17 = 20 \\
   (x, y, z) = (1, 3, 11) & \implies x + y + z = 1 + 3 + 11 = 15 \\
   (x, y, z) = (1, 4, 8) & \implies x + y + z = 1 + 4 + 8 = 13
   \end{align*}
   \]

6. The greatest possible value of \( x + y + z \) is 20.

The final answer is \(\boxed{20}\)