Problem 1485

### Part (a)

1. Base Case:
We start by verifying that the set $D_3 = \{1, 2, 3\}$ has property $D$. Indeed, we can partition $D_3$ into the triple $\{1, 2, 3\}$, where $3 = 1 + 2$.

2. Inductive Step:
Suppose $D_n = \{1, 2, \ldots, n\}$ has property $D$. We need to show that $D_{4n}$ and $D_{4n+3}$ also have property $D$.

- **For $D_{4n}$:**
We can partition $D_{4n}$ into the following sets:
$$\{1, 2n+n, 2n+n+1\}, \{3, 2n+n-1, 2n+n+2\}, \{5, 2n+n-2, 2n+n+3\}, \ldots, \{2n-1, 2n+1, 4n\}$$
Additionally, the even integers can be partitioned as:
$$\{2, 4, 6, \ldots, 2n\} = 2D_n$$
Since $D_n$ has property $D$, $2D_n$ also has property $D$. Therefore, $D_{4n}$ has property $D$.

- **For $D_{4n+3}$:**
We can partition $D_{4n+3}$ into the following sets:
$$\{1, 2n+n+2, 2n+n+3\}, \{3, 2n+n+1, 2n+n+4\}, \{5, 2n+n, 2n+n+5\}, \ldots, \{2n+1, 2n+2, 4n+3\}$$
Additionally, the even integers can be partitioned as:
$$\{2, 4, 6, \ldots, 2n\} = 2D_n$$
Since $D_n$ has property $D$, $2D_n$ also has property $D$. Therefore, $D_{4n+3}$ has property $D$.

3. Conclusion:
By induction, we have shown that if $D_n$ has property $D$, then $D_{4n}$ and $D_{4n+3}$ also have property $D$. We now check the specific cases:
- $D_3$ has property $D$.
- $3 \times 4 = 12$
- $12 \times 4 + 3 = 51$
- $51 \times 4 + 3 = 207$
- $207 \times 4 + 3 = 831$
- $831 \times 4 = 3324$

Therefore, $D_{3324}$ has property $D$.

### Part (b)

1. Sum of Elements:
The sum of the elements in $D_{3309} = \{1, 2, \ldots, 3309\}$ is:
$$\sum_{i=1}^{3309} i = \frac{3309 \times 3310}{2} = 5470095$$
This sum is odd.

2. **Property $D$ Requirement:**
For a set to have property $D$, the sum of the elements in each triple must be even. Therefore, the sum of all elements in the set must also be even.

3. Conclusion:
Since the sum of the elements in $D_{3309}$ is odd, $D_{3309}$ cannot have property $D$.