Let $n$ be a positive integer. Suppose that we have disks of radii $1, 2, . . . , n.$ Of each size there are two disks: a transparent one and an opaque one. In every disk there is a small hole in the centre, with which we can stack the
disks using a vertical stick. We want to make stacks of disks that satisfy the following conditions:
 $i)$ Of each size exactly one disk lies in the stack.
 $ii)$ If we look at the stack from directly above, we can see the edges of all of the $n$ disks in the stack. (So if there is an opaque disk in the stack,no smaller disks may lie beneath it.)
Determine the number of distinct stacks of disks satisfying these conditions.
(Two stacks are distinct if they do not use the same set of disks, or, if they do use the same set of disks and the orders in which the disks occur are different.)
To solve this problem, we need to determine the number of distinct stacks of disks that satisfy the given conditions. Let's break down the solution step by step.

1. **Understanding the Problem:**
   - We have disks of radii \(1, 2, \ldots, n\).
   - Each size has two disks: one transparent and one opaque.
   - We need to create stacks where:
     - Each size is represented by exactly one disk.
     - If an opaque disk is used, no smaller disks can be beneath it.

2. **Subset Selection:**
   - Let \(S\) be a subset of \([n] = \{1, 2, \ldots, n\}\) representing the radii of the opaque disks used in the stack.
   - If \(S\) is not empty, let \(S = \{s_1, s_2, \ldots, s_k\}\) where \(s_1 < s_2 < \ldots < s_k\) and \(k = |S|\).
   - For convenience, define \(s_0 = 0\) and \(s_{k+1} = n+1\).

3. **Arrangement of Disks:**
   - We need to arrange the disks such that all disks with radii in the interval \((s_i, s_{i+1})\) are above the disk with radius \(s_{i+1}\).
   - This means we need to arrange the disks in each interval \((s_i, s_{i+1})\) in increasing order of their radii.

4. **Counting Arrangements:**
   - The number of ways to arrange the disks in each interval \((s_i, s_{i+1})\) is given by:
     \[
     \prod_{i=0}^{k} \prod_{j=s_i+1}^{s_{i+1}-1} j = \frac{n!}{\prod_{i=1}^{k} s_i}
     \]
   - If \(S\) is an empty set, there are \(n!\) ways to stack the disks.

5. **Summing Over All Subsets:**
   - We need to sum the number of distinct stacks over all possible subsets \(S\) of \([n]\):
     \[
     n! + \sum_{S \subseteq [n], S \neq \emptyset} \frac{n!}{\prod_{s \in S} s}
     \]
   - This can be rewritten as:
     \[
     n! \left( 1 + \sum_{S \subseteq [n], S \neq \emptyset} \frac{1}{\prod_{s \in S} s} \right)
     \]

6. **Simplifying the Sum:**
   - The sum can be simplified using the product notation:
     \[
     n! \prod_{i=1}^{n} \left( 1 + \frac{1}{i} \right)
     \]
   - This product simplifies to:
     \[
     n! \prod_{i=1}^{n} \frac{i+1}{i} = n! \cdot \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{n+1}{n} = (n+1)!
     \]

Conclusion:
\[
\boxed{(n+1)!}
\]