@above we don't know if $x_2$ is positive

Purple_Planet wrote:

The solution is not correct $(x_1,x_2,\cdots, x_8)=(-3,4,-2,3,-1,1,0,2)$ works

Case I: $ x_8=0 $ If $ x_7<0, $ then $ x_5,x_3,x_1<0, $ false. So, $ x_7>0, $ which implies $ x_6,x_4,x_2<0<x_1,x_3,x_5, $ which implies, by the chain, that $ x_2<x_4<x_6<0<x_7<x_5<x_3<x_1. $ There's at most one permutation that satisfies this. Case II: $ x_7=0 $ Subcase 1: $ x_6<0 $ We have $ x_4,x_2<0<x_1,x_3,x_5 $ which implies $ x_2<x_4<x_6<0<x_5<x_3<x_1 $ and $ x_8>0. $ There are $ 4 $ possibilities of choosing $ x_8, $ so, there are at most $ 4 $ permutations that satisfy this. Subcase 2: $ x_6>0 $ $ x_5,x_3,x_1<0<x_4,x_2 $ which implies $ x_8>0 $ and $ x_1<x_3<x_5<0<x_6<x_4<x_2. $ There are $ 4 $ possibilities of choosing $ x_8, $ so, there are at most $ 4 $ permutations that satisfy this. Case III: $ x_6=0 $ If $ x_5>0, $ then $ x_4,x_2<0<x_3,x_1, $ which implies $ x_7 $ and $ x_8 $ have opposite signs, contradiction with the fact that $ x_7x_8\ge 0. $ So, $ x_5<0, $ which implies $ x_3,x_1<0<x_2,x_4, $ which implies $ x_7,x_8>0 $ and $ x_1<x_3<x_5<0<x_4<x_2. $ There are $ 12 $ ways of chosing ordered pairs $ \left( x_7,x_8 \right) $ so, there are at most $ 12 $ permutations satisfying this. Case IV: $ x_5=0 $ Either $ x_4,x_2<0, $ either $ x_3,x_1<0, $ which implies that only one of $ x_6,x_7,x_8 $ is negative, which implies that $ x_6x_7<0 $ or $ x_7x_8<0, $ false. There are no such permutations. Case V: $ x_4=0 $ If $ x_3<0, $ then $ x_1<0<x_2, $ which implies that only one of $ x_5,x_6,x_7,x_8 $ is negative, which implies that one of $ x_5x_6,x_6x_7,x_7x_8 $ is negative, contradiction. So, $ x_3>0, $ which implies that $ x_2<0<x_3,x_1 $ and that $ x_5,x_6<0 $ or $x_6,x_7<0 $ or $x_7,x_8<0, $ exclusively. That would imply that one of $ x_5x_6,x_6x_7,x_7x_8 $ is negative, contradiction. There are no permutations. Case VI: $ x_3=0 $ $ x_1 $ and $ x_2 $ have opposite signs, which implies that one of $ x_4x_5,x_5x_6,x_6x_7,x_7x_8 $ is negative, contradiction. No permutations. Case VII: $ x_2=0 $ or $ x_1=0 $ Among $ x_2,x_3,x_4,x_5,x_6,x_7,x_8 $ there are at least $ 2 $ negative numbers and at most $ 3 $ of them, which forces at least one of $ x_2x_3,x_3x_4,x_4x_5,x_5x_6,x_6x_7,x_7x_8 $ to be negative, false. No permutations. So, there are at most $ 21 $ such permutations and all $ \text{8-tuples} $ from the set $$ \{ (4,-3,3,-2,2,-1,1,0);(4,-3,3,-2,2,-1,0,1); (4,-3,3,-2,1,-1,0,2); $$$$ ;(4,-3,2,-2,1,-1,0,3) ;(3,-3,2,-2,1,-1,0,4);(-3,4,-2,3,-1,2,0,1); $$$$ ;(-3,4,-2,3,-1,1,0,2);(-3,4,-2,2,-1,1,0,3);(-3,3,-2,2,-1,1,0,4); $$$$ ;(-3,4,-2,3,-1,0,2,1);(-3,4,-2,3,-1,0,1,2);(-3,4,-2,2,-1,0,3,1); $$$$ ;(-3,4,-2,2,-1,0,1,3);(-3,4,-2,1,-1,0,3,2);(-3,4,-2,1,-1,0,2,3); $$$$ ;(-3,3,-2,2,-1,0,4,1);(-3,3,-2,2,-1,0,1,4);(-3,3,-2,1,-1,0,2,4); $$$$ ;(-3,3,-2,1,-1,0,4,2);(-3,2,-2,1,-1,0,3,4);(-3,2,-2,1,-1,0,4,3)\} , $$work. We're done. Observation: the implications are, actually equivalences, so, it's not necessary this last listing to solve the problem.