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 $x \neq y$ then $\max(x, y) \geq m$, 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
$$x = \gcd(w, x) + \gcd(m, z) = 1 + (m, z) \implies x - 1 \mid m.$$
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
$$m = \gcd(x, y) + \gcd(z, t) = \frac{m}{2} + 1 + \left(\frac{m}{2}, t\right) \implies \frac{m}{2} - 1 \mid \frac{m}{2}.$$
Which contradicts to the fact that $\frac{m}{2} > 2$. ■