$n \geq k$ -two natural numbers. $S$ -such natural, that have not less than $n$ divisors. All divisors of $S$ are written 
in descending order. What minimal number of divisors can have number from $k$-th place ?
1. Let \( S \) be a natural number with at least \( n \) divisors. We can express \( S \) in its prime factorized form as \( S = \prod_{i=1}^m p_i^{\alpha_i} \), where \( p_i \) are distinct prime numbers and \( \alpha_i \) are their respective powers.
2. The total number of divisors of \( S \) is given by \( \prod_{i=1}^m (\alpha_i + 1) \geq n \).
3. We are interested in the \( k \)-th largest divisor of \( S \). Let this divisor be \( \prod_{i=1}^m p_i^{\beta_i} \), where \( 0 \leq \beta_i \leq \alpha_i \).
4. Note that every number of the form \( \prod_{i=1}^m p_i^{\gamma_i} \), where \( \gamma_i \in [\beta_i, \alpha_i] \), must be among the \( k \) largest divisors of \( S \). Therefore, we have:
   \[
   k \geq \prod_{i=1}^m (\alpha_i - \beta_i + 1)
   \]
5. We use the inequality \( (b+1)(a-b+1) \geq a+1 \) (which follows from \( b(a-b) \geq 0 \)) to derive:
   \[
   k \prod_{i=1}^m (\beta_i + 1) \geq \prod_{i=1}^m (\alpha_i - \beta_i + 1) \prod_{i=1}^m (\beta_i + 1) = \prod_{i=1}^m (\alpha_i - \beta_i + 1)(\beta_i + 1) \geq \prod_{i=1}^m (\alpha_i + 1) = n
   \]
6. From the above inequality, we get:
   \[
   \prod_{i=1}^m (\beta_i + 1) \geq \frac{n}{k}
   \]
   Since the number of divisors must be an integer, we take the ceiling function:
   \[
   \prod_{i=1}^m (\beta_i + 1) \geq \left\lceil \frac{n}{k} \right\rceil
   \]
7. For the construction, consider \( S = p^{k-1} q^{\left\lceil \frac{n}{k} \right\rceil - 1} \), where \( q > p^{k-1} \). The \( k \)-th largest divisor of \( S \) is \( q^{\left\lceil \frac{n}{k} \right\rceil - 1} \).

The final answer is \( \boxed{ \left\lceil \frac{n}{k} \right\rceil } \).