A game is played with 16 cards laid out in a row. Each card has a black side and a red side, and initially the face-up sides of the cards alternate black and red with the leftmost card black-side-up. A move consists of taking a consecutive sequence of cards (possibly only containing 1 card) with leftmost card black-side-up and the rest of the cards red-side-up, and flipping all of these cards over. The game ends when a move can no longer be made. What is the maximum possible number of moves that can be made before the game ends? 

[i]Ray Li.[/i]

[size=85][i]See a close variant [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=810&t=500913]here[/url].[/i][/size]
1. **Initial Setup**: We start with 16 cards laid out in a row, alternating black and red, with the leftmost card being black. This can be represented as:
   \[
   B, R, B, R, B, R, B, R, B, R, B, R, B, R, B, R
   \]
   where \( B \) stands for a black card and \( R \) stands for a red card.

2. **Binary Representation**: Assign a binary digit to each card, where \( 1 \) represents a black card and \( 0 \) represents a red card. The initial sequence of cards can be represented as the binary number:
   \[
   1010101010101010_2
   \]

3. **Game Move Description**: A move consists of selecting a consecutive sequence of cards starting with a black card (1) followed by any number of red cards (0), and flipping all these cards. This means flipping a sequence like \( 10, 100, 1000, \ldots \).

4. **Effect of a Move**: Each move decreases the binary number. For example, flipping the sequence \( 10 \) (which is \( 2 \) in decimal) will decrease the binary number by \( 2 \). Similarly, flipping \( 100 \) (which is \( 4 \) in decimal) will decrease the binary number by \( 4 \), and so on.

5. **Maximum Number of Moves**: To maximize the number of moves, we should aim to decrease the binary number by the smallest possible amount in each move. The smallest possible decrease is \( 1 \), which corresponds to flipping the rightmost black card (1) and the red card (0) next to it.

6. **Achieving the Maximum Moves**: By always flipping the rightmost sequence \( 10 \), we decrease the binary number by \( 1 \) each time. Starting from \( 1010101010101010_2 \) (which is \( 43690 \) in decimal), we can make \( 43690 \) moves, each decreasing the number by \( 1 \).

7. **Conclusion**: The maximum possible number of moves is achieved by flipping the rightmost \( 10 \) sequence each time until the binary number reaches \( 0 \).

The final answer is \(\boxed{43690}\)