(本题3分)（2021$cdot$浙江$cdot$衢州市衢江区横路初级中学九年级阶段练习）将2020个边长为1的正方形按如图所示的方式排列,点$A,A_{1},A_{2},A_{3}ldotsA_{2020}$和点$M,M_{1}$,$M_{2}ldotsM_{2019}$是正方形的顶点,连接$AM_{1},AM_{2},AM_{3}ldotsAM_{2019}$分别交正方形的边$A_{1}M$,$A_{2}M_{1},A_{3}M_{2}ldotsA_{2019}M_{2018}$于点$N_{1},N_{2},N_{3}ldotsN_{2019}$,四边形$M_{l}N_{1}A_{1}A_{2}$的面积是$S_{1}$,四边形$M_{2}N_{2}A_{2}A_{3}$的面积是$S_{2},ldots$,则$S_{2019}$为$qquad$<ImageHere>
【分析】设左边第一个正方形左上角的顶点为$O$,先判定$triangleM_{l}MN_{l}backsimtriangleM_{l}OA$,利用相似三角形的性质求出$MN_{l}$的长,进而得出$S_{l}$,同理得出$S_{2}$,按照规律得出$Sn$,最后$n$取2019,计算即可得出答案.【详解】解：如图所示,设左边第一个正方形左上角的顶点为$O$<ImageHere>$because$将2019个边长为1的正方形按如图所示的方式排列$thereforeOA//MA_{l}//M_{l}A_{2}//M_{2}A_{3}//ldots//M_{2018}A_{2019}$$thereforetriangleM_{l}MN_{l}backsimtriangleM_{l}OA$$thereforefrac{MM_{1}}{OM_{1}}=frac{MN_{1}}{OA}=frac{1}{2}$$thereforeMN_{l}=frac{1}{2}$,$therefore$四边形$M_{l}N_{l}A_{l}A_{2}$的面积是$S_{l}=1-frac{1}{2}times1timesfrac{1}{2}=frac{3}{4}$;同理可得：$frac{M_{1}M_{2}}{OM_{2}}=frac{M_{1}N_{2}}{OA}=frac{1}{3}$$therefore$四边形$M_{2}N_{2}A_{2}A_{3}$的面积$S_{2}=1-frac{1}{2}times1timesfrac{1}{3}=frac{5}{6}$;$therefore$四边形$MnNnAnAn+1$的面积$Sn=1-frac{1}{2(n+1)}=frac{2n+1}{2n+2}$$thereforeS_{2019}=frac{4039}{4040}$故答案为:$frac{4039}{4040}$.【点睛】本题考查了相似三角形的判定与性质在规律型问题中的应用,数形结合并善于发现规律是解题的关键.