For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
Solution 1
We know that $n \le 30$ or else $30-n$ will be negative, resulting in a negative fraction.  We also know that $n \ge 15$ or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value.  Substituting all values $n$ from $15$ to $30$ gives us integer values for $n=15, 20, 24, 25, 27, 28, 29$.  Counting them up, we have $\boxed{\textbf{(D)}\ 7}$ possible values for $n$.  

Solution 2
Let $\frac{n}{30-n}=m$, where $m \in \mathbb{N}$. Solving for $n$, we find that $n=\frac{30m}{m+1}$. Because $m$ and $m+1$ are relatively prime, $m+1|30$. Our answer is the number of proper divisors of $2^13^15^1$, which is $(1+1)(1+1)(1+1)-1 = \boxed{\textbf{(D)}\ 7}$.