A number which when divided by $10$ leaves a remainder of $9$, when divided by $9$ leaves a remainder of $8$, by $8$ leaves a remainder of $7$, etc., down to where, when divided by $2$, it leaves a remainder of $1$, is: 
$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$
If we add $1$ to the number, it becomes divisible by $10, 9, 8, \cdots, 2, 1$. The LCM of $1$ through $10$ is $2520$, therefore the number we want to find is $2520-1=\boxed{\textbf{(D)}\ 2519}$