I have no idea about these weak bounds that appears in Tuymaada almost every year Lemma. Consider all fractions with denominator less than $n^2$ in increasing order and let $\frac{a}{b},\frac{c}{d}$ be two of the cosecutive terms. Then $ad-bc = 1$ and $b+d>n^2$ The second part of lemma is trivial since $\frac{a+c}{b+d}$ is between the two consecutive terms. The first part is a well-known fact which is easy to prove after some bash Now consider a point in $A$ and the two endpoints of the interval of $A$ which contains this point ,call it $I$.Let $a,b$ be the denominators of the endpoints of $I$. Which means , by lemma , $a+b > n^2$ and assume wlog that $a>\frac{n^2}{2}$ but on the other hand the length of $I$ is at most $\frac{1}{ab}$(again , by the lemma) and at least $\frac{2}{n^3}$ by the problem statement. Which implies that $b<n$ which will do the job for us. For each $b<n$ which is and endpoint of some interval $I$ satisfying the definition , $|I| < \frac{2}{bn^2}$ and we have at most $2b$ intervals with $b$ in the denominator of one of endpoints. So the total length $L_b$ represented to $b<n$ is atmost $L_b<\frac{4}{n^2}$ and finally: $$ the total legth \leq \sum L_b < \frac{4}{n^2} \cdot n$$Where the sume is taken over $b<n$ . Now by the unbelievable inequality $4<100$ , we're done.

Since certainly none of constants 4 and 100 is sharp, I see no reason why the solution for 4 should be considered any better than for 100, that's why I proposed it with so large constant.

Makes sense

can't this be done by diophantine approximations