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The easiest way that unfortunately takes much time is to check the first card with all numbers from 1 to 15. Using advanced guessing skills, you could easily know that the first card is neither big, in this case 11-15, nor small numbers, in this case 1-2. The reason it's not a big number is because you will need a number bigger than 15 in the third card because placing a big number makes you place a small number second and you don't have a card big enough to get the second and third card's sum to be 20. It's neither small numbers because the second and third cards have a big sum and the third and fourth cards have a huge sum. Placing a small number first causes you to place a small number in the third card as a big number is in the second card, and you need a number bigger than 15 in the fourth card.
The first card's solution to problems similar to this one tend to be around the mean of all numbers available. In this case it's 7.5, so we should start checking from 7 or 8 and then 9 and 6, etc. Let's start with 7. If the first card is seven, the second had to be 8 because 15-7=8. The third card is 12, as 20-8 shows. The fourth card is 11 since 23-12=11. The next card is 5 since 16-11=5. Using similar techniques, you easily get the sixth card is thirteen and the last card is 8. However, it's not a solution because the last card needs to be same as the second card.
Another number with good possibility for first card is 8. So we'll check it. Using techniques same as when checking if seven for the first card, you should get the following results for the seven cards in order: 8, 7, 13, 10, 6, 12, 9. Bingo. All numbers are different and less than 15, and consecutive numbers do have sums given. We'll need to continue check other numbers for first card.
If you use 9 as the first card, the 7 cards in order are 9, 6, 14, 9, etc. It's not a solution cause there's already 2 9's. Maybe 6 will work. The order of cards will be 6, 9, 11, 12, 4, 14, 7. Another solution as all are different, less than 15, and have the consecutive sums. Let's also check 5. The order would be 5, 10, 10, etc. It's obviously not the solution as there's already two 10s. Now it's time to see 4 as the first card. This time the order is 4, 11, 9, 14, 2, 16, etc. It's also not the solution because there is a need for a card bigger than 15. Finally the only possibility left is 3. The order of the 7 cards should look like this: 3, 12, 8, 15, 1, 17. It's also not the solution because it again involves a number bigger than 15.
After eliminating the obvious incorrect choices, I have done minimal guess and check to find out there are two solutions to the problem. The first one I've found had the following 7 cards in order: 8, 7, 13, 10, 6, 12, 9. The other one I've found in the process had those 7 cards in order: 6, 9, 11, 12, 4, 14, 7. In both cases the biggest cards are third, fourth, and sixth cards as I've thought. There is, I believe, other algebraic ways to do this problem involving differences.