jmc

Each group of 4 consecutive powers of $i$ adds to 0: $i+i^2+i^3+i^4=i-1-i+1=0$, $i^5+i^6+i^7+i^8=i^4(i+i^2+i^3+i^4)=1(0)=0$, and so on for positive powers of $i$. Similarly, we note that $i^{-4}=\frac{1}{i^4}=\frac{1}{1}=1$. Then $i^{-4}+i^{-3}+i^{-2}+i^{-1}=1+1\cdot i+1\cdot -1+1\cdot -i=0$, $i^{-8}+i^{-7}+i^{-6}+i^{-5}=i^{-4}(i^{-4}+i^{-3}+i^{-2}+i^{-1})=0$, and so on for negative powers of $i$. Because 100 is divisible by 4, we group the positive powers of $i$ into 25 groups with zero sum. Similarly, we group the negative powers of $i$ into 25 groups with zero sum. Therefore, $$i^{-100}+i^{-99}+\cdots +i^{99}+i^{100}=25\cdot 0+i^0+25\cdot 0=\boxed{1}$$.