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 $2(1 + 3 + \dots + 2019)$, i.e., $2 \cdot 1010^2$.