Given a positive integer $n$,  we say that a real number $x$ is $n$-good if there exist $n$ positive integers $a_1,...,a_n$ such that $$x=\frac{1}{a_1}+...+\frac{1}{a_n}.$$ Find all positive integers $k$ for which the following assertion is true:
if $a,b$ are real numbers such that the closed interval $[a,b]$ contains infinitely many $2020$-good numbers, then the interval  $[a,b]$ contains at least one $k$-good number.

(Josef Tkadlec, Czech Republic)
1. **Understanding the Problem:**
   We need to find all positive integers \( k \) such that if the interval \([a, b]\) contains infinitely many 2020-good numbers, then it must also contain at least one \( k \)-good number. A number \( x \) is \( n \)-good if it can be expressed as the sum of the reciprocals of \( n \) positive integers.

2. **Properties of \( n \)-good Numbers:**
   - Any \( n \)-good number is also an \((n+1)\)-good number. This is because if \( x = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \), then \( x \) can also be written as \( x = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} + \frac{1}{\infty} \).
   - The maximum value of an \( n \)-good number is \( n \), which occurs when all \( a_i = 1 \).

3. **Analyzing the Interval \([a, b]\):**
   - If \([a, b]\) contains infinitely many 2020-good numbers, then there must be a sequence of 2020-good numbers within this interval.
   - Consider the sequence \( x_n = 2019 + \frac{1}{n} \). Each \( x_n \) is a 2020-good number because \( x_n = 1 + 1 + \cdots + 1 + \frac{1}{n} \) (with 2019 ones and one \(\frac{1}{n}\)).

4. **Bounding \( k \):**
   - For \( k \leq 2018 \), the maximum \( k \)-good number is \( k \), which is less than 2019. Therefore, it is impossible for the interval \([a, b]\) to contain a \( k \)-good number if \( a \geq 2019 \).
   - For \( k = 2019 \), consider the sequence \( x_n = 2018 + \frac{1}{2} + \frac{1}{n} \). Each \( x_n \) is a 2020-good number because \( x_n = 1 + 1 + \cdots + 1 + \frac{1}{2} + \frac{1}{n} \) (with 2018 ones, one \(\frac{1}{2}\), and one \(\frac{1}{n}\)). This sequence converges to 2019 as \( n \to \infty \), so the interval \([a, b]\) must contain 2019.

5. **Conclusion:**
   - If \( k \geq 2019 \), then the interval \([a, b]\) containing infinitely many 2020-good numbers must also contain at least one \( k \)-good number. This is because we can always find a sequence of 2020-good numbers converging to any \( k \)-good number for \( k \geq 2019\).

The final answer is \( \boxed{ k \geq 2019 } \)