Let $(x_n)$ be a sequence of positive integers defined as follows: $x_1$ is a fixed six-digit number and for any $n \geq 1$, $x_{n+1}$ is a prime divisor of $x_n + 1$. Find $x_{19} + x_{20}$.
1. **Initial Setup and Sequence Definition:**
   Let $(x_n)$ be a sequence of positive integers where $x_1$ is a fixed six-digit number. For any $n \geq 1$, $x_{n+1}$ is a prime divisor of $x_n + 1$.

2. **Bounding the Sequence:**
   Since $x_1$ is a six-digit number, we have $x_1 \leq 999999$. Therefore, $x_1 + 1 \leq 1000000$.
   \[
   x_2 \text{ is a prime divisor of } x_1 + 1 \implies x_2 \leq \sqrt{1000000} = 1000
   \]
   Hence, $x_2 < 1000$.

3. **Continuing the Bounding Process:**
   \[
   x_2 + 1 \leq 1000 \implies x_3 \text{ is a prime divisor of } x_2 + 1 \implies x_3 \leq \sqrt{1000} \approx 31.62
   \]
   Therefore, $x_3 < 32$.

4. **Further Bounding:**
   \[
   x_3 + 1 \leq 32 \implies x_4 \text{ is a prime divisor of } x_3 + 1 \implies x_4 \leq \sqrt{32} \approx 5.66
   \]
   Hence, $x_4 < 6$.

5. **Possible Values for $x_4$:**
   Since $x_4$ is a prime number less than 6, the possible values for $x_4$ are $2, 3,$ or $5$.

6. **Analyzing Each Case:**
   - **Case 1: $x_4 = 2$**
     \[
     x_4 + 1 = 3 \implies x_5 = 3 \implies x_5 + 1 = 4 \implies x_6 = 2
     \]
     The sequence alternates between $2$ and $3$.

   - **Case 2: $x_4 = 3$**
     \[
     x_4 + 1 = 4 \implies x_5 = 2 \implies x_5 + 1 = 3 \implies x_6 = 3
     \]
     The sequence alternates between $3$ and $2$.

   - **Case 3: $x_4 = 5$**
     \[
     x_4 + 1 = 6 \implies x_5 \in \{2, 3\}
     \]
     If $x_5 = 2$, the sequence alternates between $2$ and $3$. If $x_5 = 3$, the sequence alternates between $3$ and $2$.

7. **Conclusion:**
   In all cases, the sequence alternates between $2$ and $3$ or $3$ and $2$ starting from $x_5$. Therefore, $(x_{19}, x_{20})$ will be either $(2, 3)$ or $(3, 2)$.

\[
x_{19} + x_{20} = 2 + 3 = \boxed{5}
\]