If in the number $8367542741$, 1 is added to the digits at even positions and 1 is subtracted from the digits at odd positions when counted from the left, what is the sum of the 2nd, 4th, 7th, and 9th digits of the new number formed?

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**Solution:**

1. The original number is $8367542741$.

2. Identify the digits at odd and even positions (from the left):
   - Odd positions: 1st, 3rd, 5th, 7th, 9th digits
   - Even positions: 2nd, 4th, 6th, 8th, 10th digits

3. Extract digits:
   - Positions and digits:
     - 1st: 8 (odd)
     - 2nd: 3 (even)
     - 3rd: 6 (odd)
     - 4th: 7 (even)
     - 5th: 5 (odd)
     - 6th: 4 (even)
     - 7th: 2 (odd)
     - 8th: 7 (even)
     - 9th: 4 (odd)
     - 10th: 1 (even)

4. Apply the operations:
   - Subtract 1 from digits at odd positions:
     - 1st: $8 - 1 = 7$
     - 3rd: $6 - 1 = 5$
     - 5th: $5 - 1 = 4$
     - 7th: $2 - 1 = 1$
     - 9th: $4 - 1 = 3$

- Add 1 to digits at even positions:
     - 2nd: $3 + 1 = 4$
     - 4th: $7 + 1 = 8$
     - 6th: $4 + 1 = 5$
     - 8th: $7 + 1 = 8$
     - 10th: $1 + 1 = 2$

5. The new number formed is: $7 4 5 8 4 5 1 8 3 2$

6. Find the sum of the 2nd, 4th, 7th, and 9th digits:
   - 2nd digit: $4$
   - 4th digit: $8$
   - 7th digit: $1$
   - 9th digit: $3$

7. Calculate the sum:
   $$4 + 8 + 1 + 3 = 16$$

**Answer:** $16$

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