`(x+1)/2021+(x+2)/2020=(x+2019)/3+(x+2018)/4`
`<=>(x+1+2021)/2021+(x+2+2020)/2020-(x+2019+3)/3-(x+2018+4)/4=0`
`<=>(x+2022)/2021+(x+2022)/2020-(x+2022)/3-(x+2022)/4=0`
`<=>(x+2022).(1/2021+1/2020-1/3-1/4)=0`
`<=>x+2022=0`
`<=>x=-2022`
Vậy `S={-2022}`