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. Because $259=64\cdot 4+3$, we know that if we start grouping the powers of $i$ as suggested by our first two groups above, we will have 64 groups of 4 and 3 terms left without a group: $i^{257}+i^{258}+i^{259}$. To evaluate the sum of these three terms, we use the fact that $i^{256}=(i^4{)}^{64}=1^{64}$, so $$i^{257}+i^{258}+i^{259}=i^{256}(i+i^2+i^3)=1(i-1-i)=-1.$$ So $$\begin{array}{rl}& i+i^2+i^3+\cdots +i^{258}+i^{259}\\ & =(i+i^2+i^3+i^4)+(i^5+i^6+i^7+i^8)+\cdots \\ & +(i^{253}+i^{254}+i^{255}+i^{256})+(i^{257}+i^{258}+i^{259})\\ & =0+0+\cdots +0+-1\\ & =\boxed{-1}.\end{array}$$