Round 3

Answer: $n = 2021$.
First we show that $n = 2021$ is the minimum.
Let $N = 2021$ and $m = n + N$ and suppose that the sequence $c_1, c_2, \dots, c_m$ satisfies
$$\begin{cases} c_1 + c_2 + \dots + c_l \le l, & \text{for all } 1 \le l \le m \text{ odd} \\ c_1 + c_2 + \dots + c_l \ge l, & \text{for all } 1 \le l \le m \text{ even.} \end{cases} \quad (*)$$
Then $c_1 \le 1$ and $c_2 \ge 2 - c_1 \ge 1$ and similarly, we have $c_l \le l - (c_1 + \dots + c_{l-1}) \le 1$ for any $1 \le l \le m$ odd, and $c_l \ge l - (c_1 + \dots + c_{l-1}) \ge 1$ for any $1 \le l \le m$ even.
Let $X = \{1 \le l \le m \mid c_l \le 1\}$ and $Y = \{1 \le l \le m \mid c_l > 1\}$. Then $X$ contains all $1 \le l \le m$ odd, thus $|X| \ge |Y|$. It follows that $m = |X| + |Y| \ge 2|Y|$.
Now let $\varepsilon = \frac{1}{N}$ and consider the sequence $a_1 = 1 - (N-1)\varepsilon$, $a_2 = \dots = a_N = 1 + \varepsilon$ and suppose that $b_1, b_2, \dots, b_n$ and $c_1, c_2, \dots, c_m$ are chosen to satisfy the conditions of the problem. We prove that $m \ge 2N$.
(i) Suppose that $b_k > 1$ for some $1 \le k \le n$. Then $|Y| \ge N$, therefore $m \ge 2N$.
(ii) Suppose that $b_k \le 1$ for all $1 \le k \le n$. In this case, $b_1 = b_2 = \dots = b_n = 1$ since $b_1 + \dots + b_n = n$. Thus $c_1, \dots, c_m$ is a permutation of
$$1 - (N - 1)\varepsilon, \underbrace{1 + \varepsilon, \dots, 1 + \varepsilon}_{N-1}, \underbrace{1, \dots, 1}_{n}$$
satisfying $(*)$. We claim that $c_2 = 1$. Suppose on the contrary that $c_2 = 1 + \varepsilon$. Then $c_1+c_2+c_3 \le 3$ implies that either $c_1 = 1 - (N-1)\varepsilon$ or $c_3 = 1 - (N-1)\varepsilon$. In either case, we get the contradiction $c_1 + c_2 + c_3 + c_4 \le 4 - (N-4)\varepsilon < 4$. Hence $c_2 = 1$.
It follows that $X$ contains $2$ and all $1 \le l \le m$ odd, thus $|X| \ge |Y| + 2$. Since $|Y| = N - 1$, we have $m = |X| + |Y| \ge 2N$.

This proves $n \ge N$. Now we prove $n = N$ satisfies the property of the problem. Let $0 < a_1, a_2, ..., a_N < 2$ be a sequence with $a_1 + a_2 + \dots + a_N = N$ and let $b_1 = 2 - a_1, b_2 = 2 - a_2, ..., b_N = 2 - a_N$. Then $0 < b_1, b_2, ..., b_N < 2$ and $b_1 + b_2 + \dots + b_N = N$.
Changing the indices, we may assume
$$0 < a_1 \le a_2 \le \dots \le a_k \le 1 < a_{k+1} \le \dots \le a_N < 2.$$
Then the sequence $a_1, b_1, a_2, b_2, ..., a_k, b_k, b_{k+1}, a_{k+1}, ..., b_N, a_N$ satisfies $\star$.