Estonian Math Competitions

Answer: $2\cdot 1010^2$.

We show that the least sum arises in the case of the sequence $0,1,1,3,3,5,5,\dots$ ($a_{2i}=a_{2i+1}=2i-1$ for every $i\ge 1$). Firstly, we show that this sequence meets the conditions of the problem. Indeed, if $j$ and $k$ are of the same parity then $a_{j+k}=j+k-1>(j-1)+(k-1)\ge a_j+a_k$. If $j$ and $k$ have different parities then letting, w.l.o.g., $j$ be even, we obtain $a_{j+k}=j+k-2>(j-1)+(k-2)=a_j+a_k$.

We prove now that if $a_1,a_2,a_3,\dots$ is an arbitrary sequence of integers satisfying the conditions of the problems then $a_{2n}\ge 2n-1$ if $a_{2n+1}\ge 2n-1$ for any positive integer $n$. We proceed by induction on $n$. The base cases $a_2\ge 1$ and $a_3\ge 1$ hold. If $a_{2n}\ge 2n-1$ then, from the conditions of the problem, $$a_{2n+2}\ge a_{2n}+a_2+1\ge 2n-1+1+1=2(n+1)-1,$$ which proves the induction step for even indices. The same holds for odd indices: If $a_{2n+1}\ge 2n-1$ then $$a_{2n+3}\ge a_{2n+1}+a_2+1\ge 2n-1+1+1=2(n+1)-1.$$ Thus the least sum of 2021 terms is $2(1+3+\cdots +2019)$, i.e., $2\cdot 1010^2$.