Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$. The largest integer which divides all possible numbers of the form $m^2-n^2$ is:

$ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 $
1. Let \( m \) and \( n \) be any two odd numbers with \( n < m \). We need to find the largest integer that divides all possible numbers of the form \( m^2 - n^2 \).

2. We start by expressing \( m^2 - n^2 \) using the difference of squares formula:
   \[
   m^2 - n^2 = (m - n)(m + n)
   \]

3. Since \( m \) and \( n \) are both odd, both \( m - n \) and \( m + n \) are even numbers. This is because the difference and sum of two odd numbers are always even. Therefore, \( m - n \) and \( m + n \) can be written as:
   \[
   m - n = 2a \quad \text{and} \quad m + n = 2b
   \]
   for some integers \( a \) and \( b \).

4. Substituting these into the expression for \( m^2 - n^2 \), we get:
   \[
   m^2 - n^2 = (2a)(2b) = 4ab
   \]
   This shows that \( m^2 - n^2 \) is always divisible by 4.

5. Next, we need to determine if there is a larger integer that always divides \( m^2 - n^2 \). To do this, we consider the possible values of \( m^2 \) and \( n^2 \) modulo 8. The quadratic residues modulo 8 are 0, 1, and 4. Since \( m \) and \( n \) are odd, their squares modulo 8 can only be 1:
   \[
   m^2 \equiv 1 \pmod{8} \quad \text{and} \quad n^2 \equiv 1 \pmod{8}
   \]

6. Therefore, we have:
   \[
   m^2 - n^2 \equiv 1 - 1 \equiv 0 \pmod{8}
   \]
   This means that \( m^2 - n^2 \) is always divisible by 8.

7. To check if \( m^2 - n^2 \) is divisible by 16, consider specific values of \( m \) and \( n \). For example, let \( m = 3 \) and \( n = 1 \):
   \[
   m^2 - n^2 = 3^2 - 1^2 = 9 - 1 = 8
   \]
   Here, 8 is not divisible by 16, so \( m^2 - n^2 \) is not necessarily divisible by 16.

8. Therefore, the largest integer that divides all possible values of \( m^2 - n^2 \) is 8.

The final answer is \( \boxed{8} \).