Problem 1263

1. Let $p$ and $q$ be two consecutive terms of the sequence of odd primes. By definition, both $p$ and $q$ are odd numbers.
2. The sum of two odd numbers is always even. Therefore, $p + q$ is an even number.
3. Let $r = p + q$. Since $r$ is even, it can be written as $r = 2k$ for some integer $k$.
4. To determine the number of positive divisors of $r$, we need to consider the prime factorization of $r$. Since $r$ is even, it has at least the prime factor 2.
5. The number of positive divisors of a number $n$ with prime factorization $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ is given by $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$.
6. Since $r$ is even, it has at least the divisors $1, 2, \frac{r}{2},$ and $r$. This gives us at least 4 divisors.
7. To verify, consider the smallest odd primes $p = 3$ and $q = 5$. Then $r = p + q = 3 + 5 = 8$.
8. The number 8 has the prime factorization $8 = 2^3$. The number of positive divisors of 8 is $(3 + 1) = 4$, which are $1, 2, 4,$ and $8$.

Therefore, the number of positive divisors of $p + q$ is at least 4.

The final answer is $\boxed{4}$