The sum of $n$ terms of an arithmetic progression is $153$, and the common difference is $2$. 
If the first term is an integer, and $n>1$, then the number of possible values for $n$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$
Let the progression start at $a$, have common difference $2$, and end at $a + 2(n-1)$.

The average term is $\frac{a + (a + 2(n-1))}{2}$, or $a + n - 1$.  Since the number of terms is $n$, and the sum of the terms is $153$, we have:
$n(a+n-1) = 153$
Since $n$ is a positive integer, it must be a factor of $153$.  This means $n = 1, 3, 9, 17, 51, 153$ are the only possibilities.  We are given $n>1$, leaving the other five factors.
We now must check if $a$ is an integer.  We have $a = \frac{153}{n} + 1 - n$.  If $n$ is a factor of $153$, then $\frac{153}{n}$ will be an integer.  Adding $1-n$ wil keep it an integer.
Thus, there are $5$ possible values for $n$, which is answer $\boxed{\textbf{(D)}}$.