As Mari writes down $m+3$ numbers and there are only $m$ different remainders when dividing by $m$, there must be two that give equal remainders when divided by $m$; let those numbers be $a$ and $b$. The rest of the $m+1$ include two that also give equal remainders when divided by $m$; let those be $c$ and $d$. Now $a+c$ and $b+d$ give the same remainder when divided by $m$, thus Jüri can choose the numbers $a$, $b$, $c$ and $d$.