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:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$
First, factor the [difference of squares](https://artofproblemsolving.com/wiki/index.php/Difference_of_squares).
\[(m+n)(m-n)\]
Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$, where $a$ and $b$ can be any integer.
\[(2a+2b+2)(2a-2b)\]
Factor the resulting expression.
\[4(a+b+1)(a-b)\]
If $a$ and $b$ are both even, then $a-b$ is even.  If $a$ and $b$ are both odd, then $a-b$ is even as well.  If $a$ is odd and $b$ is even (or vise versa), then $a+b+1$ is even.  Therefore, in all cases, $8$ can be divided into all numbers with the form $m^2-n^2$.
This can be confirmed by setting $m=3$ and $n=1$, making $m^2-n^2=9-1=8$.  Since $8$ is not a multiple of $3$ and is less than $16$, we can confirm that the answer is $\boxed{\textbf{(D)}}$.