jmc

To find $x_1+x_2+x_3+x_4,$ we can try to find a quartic equation whose roots are $x_1,$ $x_2,$ $x_3,$ and $x_4.$ To this end, we substitute $y=(x+1{)}^2$ into $x+4=(y-3{)}^2,$ to get $$x+4=((x+1{)}^2-3{)}^2.$$Expanding, we get $x^4+4x^3-9x=0.$ By Vieta's formulas, $x_1+x_2+x_3+x_4=-4.$

Substituting $x=(y-3{)}^2-4$ into $y=(x+1{)}^2,$ we get $$y=((y-3{)}^2-3{)}^2.$$Expanding, we get $y^4-12y^3+48y^2-73y+36=0.$ By Vieta's formulas, $y_1+y_2+y_3+y_4=12.$

Therefore, $x_1+x_2+x_3+x_4+y_1+y_2+y_3+y_4=\boxed{8}.$