smc

We first use the fact that $i^{-n}=\frac{1}{i^n}={\left(\frac{1}{i}\right)}^n=(-i{)}^n$. Note that $i^4=1$ and $(-i{)}^4=1$, so $i^n$ and $(-i{)}^n$ 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 $\boxed{3}$.