Problem 1474

Yes, there is such a set. We construct it in two stages.
Readers who are familiar with Egyptian Fractions should be aware of the fact that any rational number $0 < x < 1$ can be expressed as a sum of reciprocals of distinct integers. Indeed, the greedy algorithm delivers such an expression: at each point choose the smallest integer $s$ for which $\frac{1}{s} \le x$; this either gives equality, in which case we stop, or we start again with $x - \frac{1}{s}$ replacing $x$. We need to show this algorithm terminates and gives a sequence of distinct denominators $s$. We write the fraction in its lowest terms as $x = \frac{p}{q}$ with $1 \le p < q$. If $p = 1$ the result is immediate. Otherwise $1 < p < q$ and there exists a unique integer $s \ge 2$ such that
$$\frac{1}{s} \le x < \frac{1}{s-1}.$$
As $x - \frac{1}{s} < \frac{1}{s-1} - \frac{1}{s} = \frac{1}{s(s-1)}$, the denominator chosen in the next step must be greater than $s(s-1) \ge s$. This shows that the algorithm produces a sequence of distinct denominators.
To prove termination of this algorithm we show that the numerators are strictly decreasing. From $\frac{p}{q} < \frac{1}{s-1}$ we obtain $ps - p < q$, which is equivalent to $ps - q < p$. Therefore,
$$0 \le \frac{p}{q} - \frac{1}{s} = \frac{ps - q}{q}$$
has non-negative numerator smaller than $p$, and this remains true when we write it in lowest terms. As we cannot have an infinite strictly decreasing sequence of non-negative integers, the algorithm terminates. This completes the lemma, giving the required decomposition for rational $0 < x < 1$.

To solve the original question we wish to extend this result to rational $x \ge 1$, specifically to $x = 2020$. As the harmonic sequence diverges, for any $x \ge 1$ there is some integer $m \ge 1$ for which:
$$\sum_{j=1}^{m} \frac{1}{j} \le x < \sum_{j=1}^{m+1} \frac{1}{j}.$$
We then apply the result of the first part to
$$x - \sum_{j=1}^{m} \frac{1}{j}.$$
As this is less than $\frac{1}{1+m}$ we have distinctness and the result follows.