Let $ p$ and $ q$ be two consecutive terms of the sequence of odd primes. The number of positive divisor of $ p \plus{} q$, at least

$\textbf{(A)}\ 2  \qquad\textbf{(B)}\ 3  \qquad\textbf{(C)}\ 4  \qquad\textbf{(D)}\ 5  \qquad\textbf{(E)}\ 6$
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 \(\boxed{4}\)