How many positive integers $n$ are there such that $n!(2n+1)$ and $221$ are relatively prime?

$ 
\textbf{(A)}\ 10
\qquad\textbf{(B)}\ 11
\qquad\textbf{(C)}\ 12
\qquad\textbf{(D)}\ 13
\qquad\textbf{(E)}\ \text{None of the above}
$
1. First, we need to determine the prime factorization of 221:
   \[
   221 = 13 \times 17
   \]
   Therefore, \( n!(2n+1) \) and 221 are relatively prime if and only if \( n!(2n+1) \) is not divisible by either 13 or 17.

2. Consider the factorial part \( n! \). For \( n! \) to be relatively prime to 221, \( n \) must be less than 13 because if \( n \geq 13 \), then \( n! \) will include the factor 13, making it not relatively prime to 221.

3. Next, consider the term \( 2n+1 \). We need \( 2n+1 \) to be relatively prime to 221, which means \( 2n+1 \) should not be divisible by either 13 or 17.

4. We will check each \( n \) from 1 to 12 to see if \( 2n+1 \) is divisible by 13 or 17:
   - For \( n = 1 \):
     \[
     2(1) + 1 = 3 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 2 \):
     \[
     2(2) + 1 = 5 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 3 \):
     \[
     2(3) + 1 = 7 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 4 \):
     \[
     2(4) + 1 = 9 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 5 \):
     \[
     2(5) + 1 = 11 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 6 \):
     \[
     2(6) + 1 = 13 \quad (\text{divisible by 13})
     \]
   - For \( n = 7 \):
     \[
     2(7) + 1 = 15 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 8 \):
     \[
     2(8) + 1 = 17 \quad (\text{divisible by 17})
     \]
   - For \( n = 9 \):
     \[
     2(9) + 1 = 19 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 10 \):
     \[
     2(10) + 1 = 21 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 11 \):
     \[
     2(11) + 1 = 23 \quad (\text{not divisible by 13 or 17})
     \]
   - For \( n = 12 \):
     \[
     2(12) + 1 = 25 \quad (\text{not divisible by 13 or 17})
     \]

5. From the above calculations, the values of \( n \) that make \( n!(2n+1) \) relatively prime to 221 are:
   \[
   n = 1, 2, 3, 4, 5, 7, 9, 10, 11, 12
   \]
   This gives us a total of 10 values.

The final answer is \(\boxed{10}\)