Iranian Mathematical Olympiad

We shall show that this is impossible. Assume that there are such $1400$ numbers. First of all, note that if we divide all the numbers by $k$ then the new numbers satisfy the second condition and one of them is a divisor of $2021$, so we can assume that the greatest common divisor of the numbers is $1$.

Lemma. the gcd of each three consecutive numbers on the circle is $1$.

Proof. Assume that $a,b,c,d,e$ are five consecutive numbers on the circle, if $a,b,c$ have a common divisor $d$, then the equality $c=\gcd (a,b)+\gcd (d,e)$ implies that $d,e$ are also divisible by $d$. Similarly all the numbers should be divisible by $d$ which is in contradiction with our assumption. This completes our proof.

Assume that $m$ is the maximum number on the circle, and $x,y,m,z,t$ be five consecutive numbers. We know that $m>4$ because at least one of the numbers should be a divisor of $2021$ and $2021$ is not divisible by $2,3,4$. We have $m=\gcd (x,y)+\gcd (z,t)$ so at least one of the $\gcd (x,y)$, $\gcd (z,t)$ should be at least $\frac{m}{2}$. Without the loss of generality we can assume that $\gcd (x,y)\ge \frac{m}{2}$.

If $x\ne y$ then $\max (x,y)\ge m$, yielding $\max (x,y)=m$. If $y=m$ then by the assumption $x>\gcd (m,y)=m$ we shall arrive at a contradiction hence we have $x=m,y=\frac{m}{2}$. This implies that $\gcd (z,t)=\frac{m}{2}$ therefore $z=\frac{m}{2},y=m$ which contradicts to the fact that $y>\gcd (z,m)$.

So we have $x=y\ge \frac{m}{2}$, let $w$ be the number before $x$. From the lemma we know that $\gcd (w,x)=1$ and we have $$x=\gcd (w,x)+\gcd (m,z)=1+(m,z)⟹x-1\mid m.$$ If $m$ is odd this contradicts to the fact that $x-1\ge \frac{m}{2}-1$, if $m$ is even we have $x=\frac{m}{2}+1$ and $\gcd (m,z)=\frac{m}{2}$. But we have $$m=\gcd (x,y)+\gcd (z,t)=\frac{m}{2}+1+\left(\frac{m}{2},t\right)⟹\frac{m}{2}-1\mid \frac{m}{2}.$$ Which contradicts to the fact that $\frac{m}{2}>2$. ■