The appearance of $\sqrt{2}$ is a clue that this question will end up with a pigeon-hole principle on squares of a grid with spacing $1/20$.
We prove a more general result:

Theorem: Let $k$ be a positive integer, let $n \ge k(k+1)/2$ and let $h > 0$. Suppose that a set of points $(x_j, y_j)$ for $j = 1, 2, 3, \dots, n$ in the quarter plane is such that any two points are at a distance exceeding $h\sqrt{2}$. Then the centroid $(\bar{x}, \bar{y})$ satisfies:
$$\bar{x} + \bar{y} \ge \frac{2}{3}(k-1)h.$$
The answer to the original question follows by contradiction on setting $k = 62$ (or $k = 63$), $n = 2021$ and $h = 1/20$.

Proof of Theorem: For each point $(x_j, y_j)$ write:
$$m_j = \lfloor \frac{x_j}{h} \rfloor + \lfloor \frac{y_j}{h} \rfloor.$$
It follows then that:
$$x_j + y_j \ge h m_j.$$
By hypothesis, no two points have a distance less than or equal to $h\sqrt{2}$. Then there can be no more than one point in any $h \times h$ square. There can be at most one $j$ with $m_j = 0$, at most two with $m_j = 1$, at most three with $m_j = 2$ and so on, up to at most $k$ with $m_j = k-1$. The remaining points (at least $n - k(k+1)/2$ of them) have $m_j \ge k$.
We can deduce that the sums of the $x$'s and $y$'s is governed by the lower bound:
$$\sum_{j=1}^{n} (x_j + y_j) \ge \sum_{j=1}^{n} h m_j \ge h \sum_{i=1}^{k} i(i-1) + \left(n - \frac{k(k+1)}{2}\right) k h.$$
Recalling that:
$$\sum_{i=2}^{k} i(i-1) = 2 \sum_{i=2}^{k} \binom{i}{2} = 2 \sum_{i=2}^{k} \left\{ \binom{i+1}{3} - \binom{i}{3} \right\} = 2 \binom{k+1}{3}$$
we obtain:
$$\begin{align*} \frac{1}{h} \sum_{j=1}^{n} (x_j + y_j) &\ge \frac{(k+1)k(k-1)}{3} + \left(n - \frac{k(k+1)}{2}\right) k \\ &= \left(k - \frac{2(k-1)}{3}\right) \left(n - \frac{k(k+1)}{2}\right) + \frac{2(k-1)}{3} n \\ &\ge \frac{2(k-1)}{3} n. \end{align*}$$
Multiplying by $h$ and dividing by $n$ gives the theorem as claimed.