We first use the fact that $i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n$. Note that $i^4=1$ and $(-i)^4=1$, so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$.

For $n=0$, we have $i^0+(-i)^0=1+1=2$.
For $n=1$, we have $i^1+(-i)^1=i-i=0$.
For $n=2$, we have $i^2+(-i)^2=-1-1=-2$.
For $n=3$, we have $i^3+(-i)^3=-i+i=0$.
Hence, the answer is $\boxed{\textbf{(C)}\ 3}$.
If $S = i^n + i^{-n}$, where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is: 
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$