Starting on April 15, 2008, you can go one day backward and one day forwards to get the dates 14 and 16. If you go 15 days backward and 15 days forward, you get the dates 31 (from March) and 30 (from April). Find the least positive integer k so that if you go k days backward and k days forward you get two calendar dates that are the same.
1. **Identify the problem constraints**: We need to find the smallest positive integer \( k \) such that going \( k \) days backward and \( k \) days forward from a given date results in the same calendar date.

2. **Consider the month with the fewest days**: February is the month with the fewest days, having 28 days in a non-leap year and 29 days in a leap year.

3. **Analyze February 15 in a non-leap year**:
    - If we go 14 days backward from February 15, we reach February 1.
    - If we go 14 days forward from February 15, we reach March 1.
    - Therefore, \( k = 14 \) results in the same calendar date (1st of the month) but in different months.

4. **Verify if \( k = 14 \) is the smallest possible value**:
    - For \( k < 14 \), the dates will not match because the difference in days will not cover the transition between months.
    - For example, if \( k = 13 \):
        - Going 13 days backward from February 15 results in February 2.
        - Going 13 days forward from February 15 results in February 28.
    - These dates are not the same.

5. **Conclusion**: The least positive integer \( k \) such that going \( k \) days backward and \( k \) days forward results in the same calendar date is \( k = 14 \).

The final answer is \( \boxed{14} \).