Answering the modified question, it is if course possible to find a set $T$ of cardinality $\lfloor \frac{n}{d}\rfloor$ by just taking $\{dk|k=1,2..,\lfloor\frac{n}{d}\rfloor\}$. So we just have to show that with a set of more than that number of elements, the gcd is $1$. If not there exists a prime $p$ such that $p$ divides every element of $T$. Since it must also divide all elements of $S$, we have $p\leq d$. However there are at most $\lfloor\frac{n}{p}\rfloor$ multiples of $p$, which are all the multiples of $p$. However $|T|\leq\lfloor\frac{n}{p}\rfloor\leq\lfloor\frac{n}{d}\rfloor<1+\lfloor\frac{n}{d}\rfloor\leq |T|$, which is a contradiction.

Where can we see the whole shortlist?

Assassino9931 wrote: Where can we see the whole shortlist? Here: https://artofproblemsolving.com/community/c1512002_2019_balkan_mo_shortlist For all shortlists and olympiads see the contest collection: https://artofproblemsolving.com/community/c13_contests

GorgonMathDota wrote: Let $S \subset \{ 1, \dots, n \}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \not= 1$ and let $d$ be its smallest divisor greater than $1$. Let $T \subset \{ 1, \dots, n \}$ be a set such that $S \subset T$ and $|T| \ge 1 + \left[ \frac{n}{d} \right]$. Prove that the greatest common divisor of the elements in $T$ is $1$. [Second Version] Let $n(n \ge 1)$ be a positive integer and $U = \{ 1, \dots, n \}$. Let $S$ be a nonempty subset of $U$ and let $d (d \not= 1)$ be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the greatest common divisor of all elements of $T$ is equal to $1$. For the first version:- For what follows define $d|S$ if $d$ divides the gcd of all elements of $S$ Claim:- $d \nmid T$ Proof:- There are $ \left[ \frac{n}{d} \right]$ elements in $U$ divisible by $S$ but $T$ has $1+ \left[ \frac{n}{d} \right]$ elements,contradiction. The Claim ensures that $d \nmid T$ however $S \in T$ so $d$ is the smallest divisor of all of the elements of $S$ other than 1 but there also exists another element not divisible by $d$,hence the greatest common divisor of all elements of $T$ is equal to $1$.

Observe that $\left[ \frac{n}{d}\right]$ is the number of natural numbers such that $d| n$. Since $d$ is smallest divisor greater than $1$, $\left[\frac{n}{d}\right]$ is greater than all the factors $i>d$.This means that if $d\neq s$,$T$ can't divide other factors of $s$( Remember the subset preposition) except d since if it was the case, the inequality case would be wrong since you can't find enough numbers divisible by that number.Therefore either $gcd$ is $d$ or $1$. The preposition says that " choose total $1+\left[\frac{n}{d}\right]$ number of elements divisible by $d$, which is absurd for case $d$. Hence gcd must be $1$.