Let $p_1 = 2, p_2 = 3, p_3 = 5 ...$ be the sequence of prime numbers. Find the least positive even integer $n$ so that $p_1 + p_2 + p_3 + ... + p_n$ is not prime.
1. We need to find the least positive even integer \( n \) such that the sum of the first \( n \) prime numbers is not a prime number.
2. Let's start by calculating the sums for even values of \( n \) until we find a composite number.

   - For \( n = 2 \):
     \[
     p_1 + p_2 = 2 + 3 = 5 \quad (\text{prime})
     \]

   - For \( n = 4 \):
     \[
     p_1 + p_2 + p_3 + p_4 = 2 + 3 + 5 + 7 = 17 \quad (\text{prime})
     \]

   - For \( n = 6 \):
     \[
     p_1 + p_2 + p_3 + p_4 + p_5 + p_6 = 2 + 3 + 5 + 7 + 11 + 13 = 41 \quad (\text{prime})
     \]

   - For \( n = 8 \):
     \[
     p_1 + p_2 + p_3 + p_4 + p_5 + p_6 + p_7 + p_8 = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77 \quad (\text{composite})
     \]
     Here, \( 77 \) is composite because it is divisible by \( 7 \) and \( 11 \).

3. Since \( n = 8 \) is the smallest even integer for which the sum of the first \( n \) prime numbers is not a prime number, we conclude that \( n = 8 \).

The final answer is \( \boxed{8} \)