15 boxes are given. They all  are initially empty. By one move it is allowed to choose some boxes and to put in them numbers of apricots which are pairwise distinct powers of 2. Find the least positive integer $k$ such that it is possible
to have equal numbers of apricots in all the boxes after $k$ moves.
To solve this problem, we need to determine the minimum number of moves required to make the number of apricots in all 15 boxes equal. Each move allows us to add distinct powers of 2 to some of the boxes. 

1. **Initial Setup**:
   - We start with 15 empty boxes.
   - In the first move, we can place distinct powers of 2 in some of the boxes. For example, we can place \(2^0, 2^1, 2^2, \ldots, 2^6\) in the first 7 boxes, and leave the remaining 8 boxes empty.

2. **Subsequent Moves**:
   - In each subsequent move, we can add distinct powers of 2 to some of the boxes. The goal is to make the number of apricots in each box equal by combining these powers of 2.

3. **Combining Powers of 2**:
   - Note that combining two equal powers of 2 results in a higher power of 2:
     \[
     2^x + 2^x = 2^{x+1}
     \]
   - If we have two different powers of 2, say \(2^m\) and \(2^n\), and we want to make them equal to another pair of powers of 2, say \(2^p\) and \(2^q\), it must be that \(\{m, n\} = \{p, q\}\). This is because the sum of two distinct powers of 2 is unique.

4. **Strategy**:
   - In the first move, we place \(2^0, 2^1, 2^2, \ldots, 2^6\) in the first 7 boxes, and leave the remaining 8 boxes empty.
   - In the second move, we can add \(2^0, 2^1, 2^2, \ldots, 2^6\) to the first 7 boxes again, and place \(2^7\) in the 8th box. This results in the first 7 boxes having \(2^1, 2^2, 2^3, \ldots, 2^7\) and the 8th box having \(2^7\).
   - In the third move, we can add \(2^0, 2^1, 2^2, \ldots, 2^6\) to the first 6 boxes, and place \(2^7\) in the 9th box. This results in the first 6 boxes having \(2^2, 2^3, 2^4, \ldots, 2^7\) and the 9th box having \(2^7\).
   - We continue this process, each time making more boxes have the same number of apricots.

5. **Calculation**:
   - After the first move, we have 7 boxes with distinct powers of 2.
   - In each subsequent move, we can make at most two different numbers equal. Therefore, we need \((15 - 1) / 2 = 7\) moves to make the remaining 14 boxes equal to the first box.
   - Including the first move, we need a total of \(7 + 1 = 8\) moves.

Conclusion:
After 8 moves, all 15 boxes will have the same number of apricots.

The final answer is \(\boxed{8}\).