Note first that there is no calm representation with 2 terms: if either $a_1$ or $a_2$ is 1 or $a_1 = a_2 = \frac{1}{2}$, then the sum is greater than $\frac{17}{20}$. Otherwise, the sum is at most $\frac{1}{2} + \frac{1}{3} = \frac{5}{6} < \frac{17}{20}$, thus too small.

On the other hand, there is a representation with 3 terms, namely
$$\frac{17}{20} = \frac{1}{2} + \frac{1}{4} + \frac{1}{10},$$
showing that the smallest possible value of $k$ is 3.

Now we show that there are infinitely many calm representations: take $a_1 = 5 \cdot 2^{4n+1}$ and consider the difference
$$\frac{17}{20} - \frac{1}{5 \cdot 2^{4n+1}} = \frac{17 \cdot 2^{4n-1} - 1}{5 \cdot 2^{4n+1}}.$$
Since $2^4 = 16 \equiv 1 \mod 5$, the numerator is $17 \cdot 2^{4n-1} - 1 \equiv 17 \cdot 8 - 1 = 135 \equiv 0 \mod 5$. Thus the factor 5 cancels, and we have
$$\frac{17}{20} - \frac{1}{5 \cdot 2^{4n+1}} = \frac{A}{2^{4n+1}}$$
for some positive integer $A < 2^{4n+1}$. Since $A$ has a binary representation as $A = 2^{b_1} + 2^{b_2} + \dots + 2^{b_r}$ with distinct nonnegative integers $b_1, b_2, \dots, b_r$, we get
$$\frac{17}{20} = \frac{1}{5 \cdot 2^{4n+1}} + \frac{1}{2^{4n+1-b_1}} + \frac{1}{2^{4n+1-b_2}} + \dots + \frac{1}{2^{4n+1-b_r}},$$
which is a calm representation for every $n$. This completes the proof.