Before the math competition, Dmytro overheard Olena and Mykola talking about their birthdays.

[b]О[/b]: "The day and month of my birthday are half as large as the day and month of Mykola's birthday."
[b]М[/b]: "Also, the day of Olena's birth and the month of my birth are consecutive positive integers."
[b]О[/b]: "And the sum of all these four numbers is a multiple of $17$."

Can Dmitro determine the day and month of Olena's birth?

[i]Proposed by Olena Artemchuk and Mykola Moroz[/i]
1. **Define Variables:**
   Let \( (a, b) \) be the day and month of Olena's birthday, and \( (c, d) \) be the day and month of Mykola's birthday. According to the problem:
   - \( a = \frac{c}{2} \)
   - \( b = \frac{d}{2} \)
   - \( a \) and \( d \) are consecutive integers.
   - The sum \( a + b + c + d \) is a multiple of 17.

2. **Express Relationships:**
   From the given conditions:
   - \( a = \frac{c}{2} \implies c = 2a \)
   - \( b = \frac{d}{2} \implies d = 2b \)
   - \( a \) and \( d \) are consecutive integers, so \( d = a + 1 \) or \( d = a - 1 \).

3. **Sum Condition:**
   The sum \( a + b + c + d \) must be a multiple of 17:
   \[
   a + b + 2a + 2b = 3a + 3b
   \]
   Since \( 3a + 3b \) must be a multiple of 17, we have:
   \[
   3(a + b) = 17k \quad \text{for some integer } k
   \]
   Thus, \( a + b = \frac{17k}{3} \).

4. **Check for Integer Solutions:**
   Since \( a \) and \( b \) must be integers, \( \frac{17k}{3} \) must also be an integer. This implies \( 17k \) must be divisible by 3. The smallest \( k \) that satisfies this is \( k = 3 \), giving:
   \[
   a + b = \frac{17 \times 3}{3} = 17
   \]

5. **Possible Values for \( a \) and \( b \):**
   Since \( a \) and \( b \) are positive integers, we need to find pairs \( (a, b) \) such that \( a + b = 17 \). Possible pairs are:
   \[
   (1, 16), (2, 15), (3, 14), (4, 13), (5, 12), (6, 11), (7, 10), (8, 9), (9, 8), (10, 7), (11, 6), (12, 5), (13, 4), (14, 3), (15, 2), (16, 1)
   \]

6. **Check Consecutive Integer Condition:**
   We need \( d = 2b \) to be consecutive with \( a \). This means:
   \[
   d = a + 1 \quad \text{or} \quad d = a - 1
   \]
   Substituting \( d = 2b \) into these conditions:
   \[
   2b = a + 1 \quad \text{or} \quad 2b = a - 1
   \]

7. **Solve for Valid Pairs:**
   - For \( 2b = a + 1 \):
     \[
     a + 2b = 17 \implies a + (a + 1) = 17 \implies 2a + 1 = 17 \implies 2a = 16 \implies a = 8, b = 9
     \]
   - For \( 2b = a - 1 \):
     \[
     a + 2b = 17 \implies a + (a - 1) = 17 \implies 2a - 1 = 17 \implies 2a = 18 \implies a = 9, b = 8
     \]

8. **Verify Solutions:**
   - For \( (a, b) = (8, 9) \):
     \[
     c = 2a = 16, \quad d = 2b = 18 \quad \text{(not valid as month cannot be 18)}
     \]
   - For \( (a, b) = (9, 8) \):
     \[
     c = 2a = 18, \quad d = 2b = 16 \quad \text{(not valid as month cannot be 16)}
     \]

9. **Check Other Pairs:**
   - For \( (a, b) = (11, 6) \):
     \[
     c = 2a = 22, \quad d = 2b = 12 \quad \text{(valid as month is 12)}
     \]

Thus, the only valid solution is \( (a, b, c, d) = (11, 6, 22, 12) \).

The final answer is \( \boxed{ (11, 6, 22, 12) } \)