Nice problem. Note first that using $\textstyle \sum_{n\le i\le 2021} a_i\le ka_n$, we arrive at the expressions $(k-1)a_i\ge a_{i+1}+\cdots +a_{2021}$ for $1\le i\le 2020$. Our goal is to use this inequality together with the monotonicity to reach a lower bound, at each step, in terms of $k$ as well as $a_{2021}$. Now the main observation is as follows. This inequality itself is weaker than the pure monotonicity for $2022-k\le i\le 2020$. That is, $a_{2022-k}\ge a_{2021}$. Using this, we now have \[ (k-1)a_{2021-k}\ge a_{2022-k}+a_{2023-k}+\cdots+a_{2021}\ge ka_{2021}\implies a_{2021-k}\ge \frac{k}{k-1}a_{2021}. \]Now, \[ (k-1)a_{2020-k}\ge a_{2021-k}+a_{2022-k}+a_{2023-k}+\cdots+a_{2021} \ge \frac{k}{k-1}a_{2021} + ka_{2021}, \]implying that \[ a_{2020-k}\ge \left(\frac{k}{k-1}\right)^2 a_{2021}. \]Iterating in the exact same way, we find \[ a_1\ge \left(\frac{k}{k-1}\right)^{2021-k}a_{2021}\implies a_{2021}\le \left(1-\frac1k\right)^{2021-k}a_1. \]With this, it now suffices to show \[ 4\left(1-\frac1k\right)^k\ge 1. \]To that end, we study the function \[ \varphi(x):=\left(1-\frac1x\right)^x,\qquad x\in[2,\infty). \]Check that for $g(x)=\ln \varphi(x)$, \[ g'(x) = \ln\left(1-\frac1x\right)+\frac{1}{x-1} \ge 0, \]using $\ln (1+t)\ge t/(1+t)$ valid for all $t>-1$. Consequently, $g'(\cdot)\ge 0$, and $g$ is increasing. Since $\ln$ is monotonic, it follows $\varphi$ is increasing too. That is, $\varphi(x) \ge \varphi(2)=0.25$. This concludes the proof.