The correct answer is option (B):
7
Let the arithmetic progression be denoted by a, a+d, a+2d, ..., a+(n-1)d.
We are given that the arithmetic progression starts with 1 and ends with 1000. So, a = 1 and a+(n-1)d = 1000.
Since a = 1, we have 1 + (n-1)d = 1000. Thus, (n-1)d = 999.
We are looking for arithmetic progressions with at least 3 elements, so n >= 3.
This means n-1 >= 2.
Since (n-1)d = 999, we need to find the number of divisors of 999 that are at least 2.
The prime factorization of 999 is 3^3 * 37.
The divisors of 999 are 1, 3, 9, 27, 37, 111, 333, 999.
Since n-1 can take these divisor values (n-1 >= 2), we have:
n-1 = 3, d = 333, n = 4. The progression is 1, 334, 667, 1000.
n-1 = 9, d = 111, n = 10. The progression is 1, 112, 223, ..., 1000.
n-1 = 27, d = 37, n = 28.
n-1 = 37, d = 27, n = 38.
n-1 = 111, d = 9, n = 112.
n-1 = 333, d = 3, n = 334.
n-1 = 999, d = 1, n = 1000.
The possible values for n-1 are 3, 9, 27, 37, 111, 333, and 999. There are 7 such values.
Each value of n-1 leads to a valid arithmetic progression with at least 3 elements (n >= 3), because d = 999 / (n-1) is an integer and the elements are in the set S. The largest element in the arithmetic progression will be 1000, which is also in S.
Therefore, there are 7 such arithmetic progressions.
Final Answer: The final answer is $\boxed{7}$