The sequence $(a_n)$ is defined recursively by $a_0=1$, $a_1=\sqrt[19]{2}$, and $a_n=a_{n-1}a_{n-2}^2$ for $n \ge 2$. What is the smallest positive integer $k$ such that the product $a_1a_2 \cdots a_k$ is an integer?

$\textbf{(A)}\ 17 \qquad
\textbf{(B)}\ 18 \qquad
\textbf{(C)}\ 19 \qquad
\textbf{(D)}\ 20 \qquad
\textbf{(E)}\ 21$
1. Define \( x_n = \log_2(a_n) \). Given the recursive relation \( a_n = a_{n-1} a_{n-2}^2 \), we can express this in terms of \( x_n \) as:
   \[
   x_n = x_{n-1} + 2x_{n-2}
   \]
   with initial conditions \( x_0 = \log_2(1) = 0 \) and \( x_1 = \log_2(\sqrt[19]{2}) = \frac{1}{19} \).

2. The characteristic equation for the recurrence relation \( x_n = x_{n-1} + 2x_{n-2} \) is:
   \[
   \lambda^2 - \lambda - 2 = 0
   \]
   Solving this quadratic equation, we get the roots:
   \[
   \lambda = 2 \quad \text{and} \quad \lambda = -1
   \]

3. Therefore, the general solution for \( x_n \) is:
   \[
   x_n = A \cdot 2^n + B \cdot (-1)^n
   \]
   Using the initial conditions to solve for \( A \) and \( B \):
   \[
   x_0 = 0 \implies A + B = 0 \implies B = -A
   \]
   \[
   x_1 = \frac{1}{19} \implies 2A - A = \frac{1}{19} \implies A = \frac{1}{57}
   \]
   Thus, \( B = -\frac{1}{57} \).

4. Substituting \( A \) and \( B \) back into the general solution, we get:
   \[
   x_n = \frac{1}{57} \cdot 2^n - \frac{1}{57} \cdot (-1)^n
   \]

5. To find the smallest positive integer \( k \) such that the product \( a_1 a_2 \cdots a_k \) is an integer, we need \( x_1 + x_2 + \cdots + x_k \) to be an integer. Summing \( x_n \) from 1 to \( k \):
   \[
   \sum_{i=1}^k x_i = \sum_{i=1}^k \left( \frac{1}{57} \cdot 2^i - \frac{1}{57} \cdot (-1)^i \right)
   \]
   \[
   = \frac{1}{57} \left( \sum_{i=1}^k 2^i - \sum_{i=1}^k (-1)^i \right)
   \]

6. Evaluate the sums separately:
   \[
   \sum_{i=1}^k 2^i = 2 + 2^2 + \cdots + 2^k = 2(2^k - 1)
   \]
   \[
   \sum_{i=1}^k (-1)^i = \begin{cases} 
   0 & \text{if } k \text{ is even} \\
   -1 & \text{if } k \text{ is odd}
   \end{cases}
   \]

7. Combining these results:
   \[
   \sum_{i=1}^k x_i = \frac{1}{57} \left( 2(2^k - 1) - \sum_{i=1}^k (-1)^i \right)
   \]
   For \( k = 17 \) (an odd number):
   \[
   \sum_{i=1}^k (-1)^i = -1
   \]
   \[
   \sum_{i=1}^k x_i = \frac{1}{57} \left( 2(2^{17} - 1) + 1 \right) = \frac{1}{57} \left( 2^{18} - 2 + 1 \right) = \frac{1}{57} (2^{18} - 1)
   \]

8. Since \( 2^{18} - 1 \) is divisible by 57 (as shown in the solution), \( \frac{1}{57} (2^{18} - 1) \) is an integer.

Therefore, the smallest positive integer \( k \) such that the product \( a_1 a_2 \cdots a_k \) is an integer is \( \boxed{17} \).