Estonian Math Competitions

Answer: $1,908,1011,1918,2021$ is the only possibility.

Let $(a,1919-a)$ and $(b,1919-b)$ be the two pairs of numbers with sum $1919$. If these sums had a common addend then both addends would be the same whence the choices of two numbers would not be different. Thus $a,b,1919-a$ and $1919-b$ are pairwise distinct. Let $x$ be the fifth chosen number; then the total sum of chosen numbers is $3838+x$.

Similarly, we see that the five numbers must be representable as $c,2929-c,d,2929-d,y$; their total sum is $5858+y$. From $3838+x=5858+y$, one gets $x-y=2020$. The only possibility for satisfying the latter equality is $x=2021,y=1$. Hence one of $a,b,1919-a,1919-b$ equals $1$ and one of $c,d,2929-c,2929-d$ is $2021$; w.l.o.g., $a=1$ and $c=2021$. Then the set of chosen numbers contains also $1919-1=1918$ and $2929-2021=908$. W.l.o.g., $b=908$. The last chosen number must be $1919-908=1011$.

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Alternative solution.

Let the chosen numbers be $a,b,c,d,e$. W.l.o.g., $a+b=1919$. If the other way to obtain $1919$ as the sum of two chosen numbers used either $a$ or $b$ as addend then the other addend would have to be $b$ or $a$, respectively, which is not allowed. Hence, w.l.o.g., $c+d=1919$.

Similarly, one must use four distinct numbers to get $2929$ as the sum of two chosen numbers in two different ways. If these four numbers were $a,b,c,d$, then $a+b=c+d=1919$ implies $a+b+c+d=3838$ while also $a+b+c+d=2929+2929=5858$. The contradiction shows that one must use the number $e$ in at least one pair of numbers with sum $2929$. W.l.o.g., $d+e=2929$. The other pair of numbers with sum $2929$ must use exactly two numbers among $a,b,c$, but $a+b=1919$. Hence, w.l.o.g., $b+c=2929$.

These equalities together imply $b=1919-a$, $c=2929-b=1010+a$, $d=1919-c=909-a$ and $e=2929-d=2020+a$. The latter implies $a=1$ and $e=2021$. Substituting $a=1$ into the other equalities gives $b=1918$, $c=1011$ and $d=908$.