Let $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, $(x_4, y_4)$, and $(x_5, y_5)$ be the vertices of a regular pentagon centered at $(0, 0)$. Compute the product of all positive integers k such that the equality $x_1^k+x_2^k+x_3^k+x_4^k+x_5^k=y_1^k+y_2^k+y_3^k+y_4^k+y_5^k$ must hold for all possible choices of the pentagon.
1. **Vertices on the Unit Circle:**
   Since the vertices of the pentagon lie on the unit circle centered at \((0,0)\), we can represent them using complex numbers. Let the vertices be \(z_1, z_2, z_3, z_4, z_5\) where \(z_i = \exp\left(\frac{2\pi i (i-1)}{5}\right)\) for \(i = 1, 2, 3, 4, 5\). Here, \(\exp\left(\frac{2\pi i}{5}\right)\) is a primitive 5th root of unity, denoted by \(\omega\).

2. **Sum of Powers of Vertices:**
   We need to compute the product of all positive integers \(k\) such that the equality \(x_1^k + x_2^k + x_3^k + x_4^k + x_5^k = y_1^k + y_2^k + y_3^k + y_4^k + y_5^k\) holds for all possible choices of the pentagon. Since the vertices are on the unit circle, we have:
   \[
   z_i = \exp\left(\frac{2\pi i (i-1)}{5}\right)
   \]
   and thus:
   \[
   z_i^k = \exp\left(\frac{2\pi i k (i-1)}{5}\right)
   \]

3. **Sum of \(k\)-th Powers:**
   The sum of the \(k\)-th powers of the vertices is:
   \[
   \sum_{i=1}^5 z_i^k = \sum_{i=1}^5 \exp\left(\frac{2\pi i k (i-1)}{5}\right)
   \]
   This sum is zero unless \(k\) is a multiple of 5, due to the properties of the roots of unity. Specifically, the sum of the \(k\)-th powers of the roots of unity is zero unless \(k\) is a multiple of the order of the roots (which is 5 in this case).

4. **Equality Condition:**
   For the equality \(x_1^k + x_2^k + x_3^k + x_4^k + x_5^k = y_1^k + y_2^k + y_3^k + y_4^k + y_5^k\) to hold for all possible choices of the pentagon, the sum of the \(k\)-th powers of the \(x\)-coordinates must equal the sum of the \(k\)-th powers of the \(y\)-coordinates. Since the vertices are symmetric and lie on the unit circle, this condition is satisfied if and only if \(k\) is a multiple of 5.

5. **Product of Positive Integers \(k\):**
   The positive integers \(k\) that satisfy the condition are the multiples of 5. The smallest such integers are \(5, 10, 15, 20, \ldots\). However, we are interested in the product of all such \(k\) less than 10. The only such \(k\) is 5.

The final answer is \(\boxed{5}\)