Problem 1917

Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \subset T$, we immediately get that $t/s$. Let us assume for the sake of contradiction that $t \neq 1$. From the previous observation we get that $t \geq d$.
By taking into account that $|T| \geq 1 + \lfloor \frac{n}{d} \rfloor$, we infer that we can find at least $1 + \lfloor \frac{n}{d} \rfloor$ elements in $T$. All of them will be divisible by $t$, and the largest of them, which we shall denote by $M$, will be at least $t \cdot (1 + \lfloor \frac{n}{d} \rfloor)$. On the other hand, $t \geq d$, hence
$$M \geq t \cdot (1 + \lfloor \frac{n}{d} \rfloor) \geq d \cdot (1 + \lfloor \frac{n}{d} \rfloor) > d \cdot \frac{n}{d} = n.$$
Therefore, $M > n$, which contradicts the fact that $M \in \{1, \dots, n\}$.
In conclusion, $t = 1$, as desired. $\square$

Solution:
We will show that $k_{\min} = 1 + \lfloor \frac{n}{d} \rfloor$ (here $\lfloor \cdot \rfloor$ denotes the integer part).
Obviously, the number of elements of $S$ is not greater than $\lfloor \frac{n}{d} \rfloor$, i.e. $|S| \leq \lfloor \frac{n}{d} \rfloor$, and $S \neq U$.
If $S \subset T$ and the greatest common divisor of elements of $T$ is equal to $1$, then $|T| \geq |S| + 1$.
1) Assume that $|S| < \lfloor \frac{n}{d} \rfloor$. Let $T$ be the subset of $U$, consisting of all multiples of $d$ in $U$. Thus, $|T| = \lfloor \frac{n}{d} \rfloor$ and $S \subset T$. Therefore, the greatest common divisor of all elements of $T$ is $d > 1$. Thus, $k \geq 1 + \lfloor \frac{n}{d} \rfloor$.
2) Assume $|S| = \lfloor \frac{n}{d} \rfloor$. Let $T$ be any subset of $U$ with $S \subset T, S \neq T$. Therefore, $|T| \geq 1 + \lfloor \frac{n}{d} \rfloor$. Let $q$ be the greatest common divisor of all elements of $T$. Assume that $q > 1$. Therefore, $q$ is a common divisor of all elements of $S$ as well. Hence, $q \geq d$. It follows that $|T| \leq \lfloor \frac{n}{q} \rfloor \leq \lfloor \frac{n}{d} \rfloor$, contradiction. Hence, $q = 1$.
Therefore, the minimal possible value of $k$ is $1 + \lfloor \frac{n}{d} \rfloor$. $\square$