Answer: only $a_{i}=i$. Denote $A=m-n$. Looking at pairs $(m,n)$ with difference greater that some $C$ is equivalent to looking at pairs $(A,n)$ with $A$ greater than $C$. If $a_{i}=i$ then $GCD(m+i,n+a_{i})=GCD(m-n,n+i)=GCD(A,n+i)$. Suppose $\exists i,j : GCD(A,n+i), GCD(A,n+j) > \dfrac{A}{2020}$. Then $GCD(A,n+i) = \dfrac{A}{p} > \dfrac{A}{2020} \Rightarrow p<2020$. Analogously $GCD(A,n+j) = \dfrac{A}{q}$ and $q<2020$. Denote $x=n+i$ and $y=n+j$. We have $x = u \cdot \dfrac{A}{p};\; y = v \cdot \dfrac{A}{q}$ and $|x-y|<2021$. Therefore $A > 20^{21} > 2021 \cdot 2020 \cdot 2020 > pq|x-y| = |A(qu-pv)| \ge A$. Therefore not more than one term is greater than $\dfrac{A}{2020}$. Therefore $\sum \le \text{max}GCD + 2020 \cdot \dfrac{A}{2020} \le A + A = 2A$. Therefore $a_{i}=i$ satisfies the condition. Now suppose there exists some other permutation that satisfies the condition. Denote $r_{i} = i - a_{i}$. Note that $\sum r_{i} = 0$, $|r_{i}| < 2020$ and there exists some nonzero $r_{i}$. Therefore the greatest $r_{i}$ is positive. Clearly all $r_{i}$ can't be equal. Therefore we can pick $i,j$ such that $r_{i}$ is the greatest and $r_{i} \ne r_{j}$. Then $2019 + r_{i} + r_{j} = (2019 + r_{j}) + r_{i} \ge r_{i} > 0$. Note that $\forall k\; GCD(m+k, n+a_{k}) = GCD(A + r_{k}, n + a_{k})$. I claim that there exist some $A$ and $n$ such that $A + r_{i} | n+a_{i}$ and $A + r_{j} | n+a_{j}$. We first pick $A$ such that $GCD(A + r_{i}, A + r_{j}) = GCD(r_{i} - r_{j}, A + r_{j}) = 1$. Consider some prime $q$ which divides $r_{i} - r_{j}$. Then consider the remainders $0,1,\dots,q-1$ and pick any remainder $t_{q}$ other that $r_{j}$. Then by Chinese remainder theorem (CRT) there exists $A$ such that $A \equiv t_{q} (mod \; q)$. Then $q \nmid A + r_{j}$ and $GCD(r_{i} - r_{j}, A + r_{j}) = 1$. Now if $A$ is too small we add $Q = \prod q$ until $A$ is greater than $20^{21}$. Now pick $n$ such that $n \equiv -r_{i} (mod \; A + r_{i})$ and $n \equiv -r_{j} (mod \; A + r_{j})$. We can do that by CRT since $A + r_{i}$ and $A + r_{j}$ are coprime. Then $GCD(A + r_{i}, n + a_{i}) + GCD(A + r_{j}, n + a_{j}) = 2A + r_{i} + r_{j}$ and $2A > \sum \ge 2A + r_{i} + r_{j} + 2019 > 2A$. Contradiction

Lovely problem! The only solution is $$\boxed{a_i=i \ \ \text{for } i=1,2,\dots 2021}$$ Let $<a_i>$ be a permutation that satisfies the condition. Define $b_i=a_i-i$ for $i=1,2,\dots 2021$. Choose an $i$ with the maximum $b_i$. Note that $b_i \geq 0$. Suppose $b_i \geq 1$. Then there exists a $j$ such that $b_j<b_i$. Choose a positive integer $d>20^{2100}$ such that $d+b_i$ is a prime number. Now, $$\gcd (d+b_i,d+b_j)=\gcd (d+b_i,b_j-b_i) =1$$so we can choose an $m>0$ such that $d+b_i \mid m+i$ and $d+b_j \mid m+j$. Then put $n=m+d$ to get $$2d-1 \geq \sum_{k=1}^{2021} \gcd(m+k,n+a_k) \geq 2019+ \gcd(m+i,d+b_i)+\gcd(m+j,d+b_j) =2019+2d+b_i+b_j$$$$\implies b_j \leq -2020-b_i \leq -2021$$which is impossible because $$b_j=a_j-j \geq 1-2021=-2020$$Therefore $b_i=0$ $\implies$ $b_j=0$ for all $j$ by maximality of $i$ and the fact that sum of all $b_j$ is $0$ $\implies$ $a_j=j$ for $j=1,2,\dots 2021$ Now we prove that the identity permutation indeed satisfies the condition. Since the condition is now symmetric, WLOG $m>n$. Let $m-n=d$. It is sufficient to prove that $$\sum_{k=1}^{2021} \gcd(d,n+k) < 2d$$for all $n>0$ and $d>20^{21}$. Define $$c_i=\frac{d}{\gcd(d,n+i)}$$for $i=1,2,\dots 2021$. $d \mid c_i(n+i)$ and $d \mid c_j(n+j)$ $\implies$ $d \mid c_i c_j (i-j)$ for all $i,j$. $$\implies c_ic_j \geq \frac{d}{|i-j|} \geq \frac{d}{2020}$$for all distinct $i,j$. This means that there is at most one $c_i$ which is strictly less than $\sqrt{\frac{d}{2020}}$. Therefore, $$\sum_{k=1}^{2021} \gcd(d,n+k) = d \sum_{k=1}^{2021} \frac{1}{c_k} \leq d \left(1 + 2020 \sqrt{\frac{2020}{d}} \right) <2d$$because $20^{21}>2020^3$. $\blacksquare$

Wow !! This problem is generalization of https://artofproblemsolving.com/community/c6h1770752p11619138 INMO 2019..

MatBoy-123 wrote: Wow !! This problem is generalization of https://artofproblemsolving.com/community/c6h1770752p11619138 INMO 2019.. Exactly, I just generalized INMO 2019 problem and proposed it on ARMO.

Assume that some $\pi\neq \text{id}$ satisfies the given inequality for all $m$ and $n$. Then, $\pi$ forms a cycle of size at least two, so let $a$ and $k\geq 1$ such that \[a\to\pi(a)\to\pi^2(a)\to\cdots\to\pi^k(a)\to a.\]Furthermore, let $n=m+C$ for some $C>20^{21}.$ Since $\pi$ satisfies the given, it follows that for all $m$ and $C>20^{21}$ we have\[S:=\sum_{i=0}^k\gcd(m+\pi^i(a),m+C+\pi^{i+1}(a))<2C.\]Now, assume that $\pi^{i+1}(a)-\pi^i(a)\neq\pi^{i+2}(a)-\pi^{i+1}(a)$ for some $i.$ Then, we can choose $C>10^{21}$ such that \[\gcd(C+\pi^{i+1}(a)-\pi^i(a),C+\pi^{i+2}(a)-\pi^{i+1}(a))=1\]and therefore, for that value of $C$ by the Chinese Remainder Theorem, we can choose $m$ such that \begin{align*}m &\equiv -\pi^i(a)\bmod{C+\pi^{i+1}(a)-\pi^i(a)} \\ &\equiv -\pi^{i+1}(a)\bmod{C+\pi^{i+2}(a)-\pi^{i+1}(a)}\end{align*}and for this pair $(m,C),$ using the fact that $\gcd(x,y)=\gcd(x,|x-y|)$ we get that \begin{align*}S &\geq\gcd(m+\pi^i(a),m+C+\pi^{i+1}(a))+\gcd(m+\pi^{i+1}(a),m+C+\pi^{i+2}(a)) \\ &=(C+\pi^{i+1}(a)-\pi^i(a))+(C+\pi^{i+2}(a)-\pi^{i+1}(a))=2C+\pi^{i+2}(a)-\pi^i(a)\end{align*}or, in other words $S\geq 2C+\pi^{i+2}(a)-\pi^i(a).$ But $S<2C$ so $\pi^{i+2}(a)-\pi^i(a)<0.$ But we can do the same reasoning by taking $m=n+C$ and for some proper $n$ and $C$ we'd get that $\pi^{i+2}(a)-\pi^i(a)>0.$ So, if there exists some $i$ such that $\pi^{i+1}(a)-\pi^i(a)\neq\pi^{i+2}(a)-\pi^{i+1}(a)$ then we get that $0>\pi^{i+2}(a)-\pi^i(a)>0,$ contradiction. Assume such an $i$ doesn't exist. Then, $\pi^{i+1}(a)-\pi^i(a)=c$ for all $i$ so by summing \[0=\sum_{i=0}^k\pi^{i+1}(a)-\pi^i(a)=\sum_{i=0}^kc=(k+1)c\]so $c=0.$ But then, for all $i$ we get that $\pi^i(a)=\pi^{i+1}(a)$ so $a\to\pi(a)\to\cdots\to\pi^k(a)$ isn't actually a cycle, contradiction. Therefore, all $\pi\neq\text{id}$ fail to satisfy the given inequality. Now, it remains to show that $\pi=\text{id}$ works. Observe that \[\sum_{i=1}^{2021}\gcd(n+i,m+i)=\sum_{i=1}^{2021}\gcd(n+i,C).\]However, there do not exist $i$ and $j$ such that $\gcd(n+i,C)=C/a$ and $\gcd(n+j,C)=C/b$ are both greater than $C/2020^2$ because \begin{align*}2020&\geq|i-j|=|n+i-n+j|\geq\gcd(\gcd(n+i,C),\gcd(n+j,C)) \\ &=\gcd(C/a,C/b)=C/\text{lcm}(a,b)\geq C/2020^4\end{align*}and since $C>20^{21}$ we clearly cannot have $2020\geq C/2020^4.$ Therefore \[\sum_{i=1}^{2021}\gcd(n+i,C)\leq C+2020\cdot\frac{C}{2020^2}<2C.\]

Wow, this is very nice! The answer is only $a_i = i$ for all $i = 1,2,\cdots,2021$. Without loss of generality assume $n > m$ as otherwise just consider the inverse permutation and let $d = n-m$. If $a_i = i$ for all $i$, then the problem is equivalent to showing that $\gcd(d,n+1) + \gcd(d,n+2) + \cdots + \gcd(d,n+2021) < 2d$. Suppose not, then there is some index $i$ such that $\gcd(d,n+i) \geqslant \frac{2d}{2021}$. Consider another index $j \neq i$. Then we have$$\gcd(d,n+j) \leqslant \gcd(\gcd(d,n+j), \gcd(d,n+i)) \cdot \frac{n+i}{\gcd(d,n+i)} \leqslant |j-i| \cdot \frac{2021}{2} \leqslant 2021^2$$where the inequality is essentially due to checking the common part of $\gcd(d,n+j)$ with $\gcd(d,n+i)$ multiplied by the "leftover" part. This implies that $\sum_{k=1}^{2021} \gcd(d,n+k) \leqslant d + 2020 \cdot 2021^2 < 2d$ as $d > 20^{21}$ and $\gcd(d,n+i) \leqslant d$, as desired. Now assume that the permutation is not the identity and consider all possible values of $a_i - i$. There must exist indices $k, \ell$ such that $a_k - k, a_{\ell} - \ell \neq 0$ with both values distinct and $a_{k} + a_{\ell} - k - \ell \geqslant 0$. Let $a = a_k - k$ and $b = a_{\ell} - \ell$. Then we know that the left-hand side is at least $\gcd(m+k,n+a_k) + \gcd(m+\ell, n+a_{\ell}) = \gcd(d+a,m+k) + \gcd(d+b,m+\ell)$. Choose $d$ to be sufficiently large and not divisible by $|a-b|$. This ensures that $d+a$ and $d+b$ are coprime. Now, by CRT choose $m$ to satisfy $m \equiv -k \pmod{d+a}$ and $m \equiv -\ell \pmod{d+b}$, possible since they are coprime. This means that the left-hand side is at least $(d+a)+(d+b) = 2d+a+b \geqslant 2d$, and so the permutation does not work for all values of $m$ and $n$, as desired. $\blacksquare$