This question is really boring, All you have to do is show a construction to get $1$

Yep, but construction is quite intiutive: $(1-2)+(3-4)+\cdots +(1009-1010)+(1012-1011)+(1014-1013)+\cdots +(2022-2021)=-1-1-\cdots -1+1+\cdots +1=-505+506=1$ and $\pmod{2}$ shows $0$ is not attainable. @below in senior P1 it was mentioned non-negative. That's why I noted this also. But yes, for this one $0$ is not needed.

Jalil_Huseynov wrote: Yep, but construction is quite intiutive: $(1-2)+(3-4)+\cdots +(1009-1010)+(1012-1011)+(1014-1013)+\cdots +(2022-2021)=-1-1-\cdots -1+1+\cdots +1=-505+506=1$ and $\pmod{2}$ shows $0$ is not attainable. Iora wrote: Find the minimum positive value of $ 1*2*3*4*...*2020*2021*2022$ where you can replace $*$ as $+$ or $-$

Iora wrote: Find the minimum positive value of $ 1*2*3*4*...*2020*2021*2022$ where you can replace $*$ as $+$ or $-$ Another solution: 1-2-3+4=0, 5-6-7+8=0,.....,2017-2018-2019+2020=0 and 2021-2020=1 So adding those up we get the answer 1.

Another construction $(1-2)+(2-3)... (2021-2022)$ adding 1012 back gives 1

thats so weird and easy question for junior olympiad

my construction was: $1-2+3-4+..........+(1009-1010)+1012-1011+.... 2022-2021$

Such problems, we can do something to get zero then adding the last number of the sequence. I think it is much better; (1-2)+(3-4)+(5-6)+...+(1007-1008)+(1009-1010)+(1012-1011)+(1014-1013)+...(2022-2021)

$(1+2022)-(2+2021)+(3+2020)-....+(1004+1009)-(1005+1008)+1007-1006=1$ Wow! So hard

Iora wrote: Find the minimum positive value of $ 1*2*3*4*...*2020*2021*2022$ where you can replace $*$ as $+$ or $-$ since we are given a positive integer we can try to get 1 and we can; $(1-2)+(4-3)=0$ $(5-6)+(8-7)=0$ … $(2017-2018)+(2020-2019)=0$ $(2022-2021)=1$ summing these up would give the desired result which is $1$