If it is possible to find six elements, whose sum are divisible by $6$, from every set with $n$ elements, what is the least $n$ ?

$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 9$
To solve this problem, we need to determine the smallest number \( n \) such that any set of \( n \) integers contains a subset of six elements whose sum is divisible by 6. We will use the Erdős–Ginzburg–Ziv theorem, which states that for any \( 2n-1 \) integers, there exists a subset of \( n \) integers whose sum is divisible by \( n \).

1. **Understanding the Erdős–Ginzburg–Ziv theorem**:
   The theorem states that for any \( 2n-1 \) integers, there exists a subset of \( n \) integers whose sum is divisible by \( n \). For our problem, we need to find the smallest \( n \) such that any set of \( n \) integers contains a subset of 6 elements whose sum is divisible by 6.

2. **Applying the theorem**:
   According to the theorem, for \( n = 6 \), we need \( 2 \cdot 6 - 1 = 11 \) integers to guarantee that there is a subset of 6 integers whose sum is divisible by 6. Therefore, \( f(6) = 11 \).

3. **Verification**:
   We need to verify that 11 is indeed the smallest number. If we have fewer than 11 integers, say 10, it is possible to construct a set of 10 integers such that no subset of 6 integers has a sum divisible by 6. For example, consider the set \( \{0, 0, 0, 0, 0, 1, 1, 1, 1, 1\} \). Any subset of 6 integers from this set will have a sum that is not divisible by 6.

4. **Conclusion**:
   Since 11 is the smallest number of integers required to ensure that any subset of 6 integers has a sum divisible by 6, the least \( n \) is 11.

The final answer is \( \boxed{11} \).