Here's my solution : $Denote~11~numbers~a1,a2,.....a11.~We~assume~they~sum~to~100.~Then~the~12th~number~must~be~(a1)+1~the~next~number~must~be~a2-1~then~a3+1~and~so~on~to~a11+1~the~next~number~must~be~a1+1 ~or~a1~if~we~want~the~sequence~to~sum~to~100~or~101.~But~both~these~terms~have~already~come~in~the~sequence~so~our~answer~is~22.$

Here is an example that shows that $22$ is the best we can do: $$(-90, -70, -50, -30, -10, 10, 30, 50, 70, 90, 100, -89, -71, -49, -31, -9, 9, 31, 49, 71, 89, 101)$$