What is the greatest number of integers that can be selected from a set  of $2015$ consecutive numbers so that no sum of any two selected numbers is divisible by their difference?
1. **Assume we have chosen two integers \( a \) and \( b \) from the set of 2015 consecutive numbers, where \( a > b \).**

2. **Consider the difference \( a - b \). We need to ensure that \( a + b \) is not divisible by \( a - b \).**

3. **First, check the case when \( a - b = 1 \):**
   \[
   a - b = 1 \implies a + b = 2b + 1
   \]
   Since \( a - b = 1 \), it must divide \( a + b \). However, \( 1 \) divides any integer, so this does not provide a contradiction. Therefore, we need to consider larger differences.

4. **Next, check the case when \( a - b = 2 \):**
   \[
   a - b = 2 \implies a + b = 2b + 2
   \]
   Since \( a - b = 2 \), it must divide \( a + b \). However, \( 2 \) divides \( 2b + 2 \), which is always true. This is a contradiction.

5. **Thus, we conclude that \( a - b \geq 3 \).**

6. **Let \( x_1, x_2, \ldots, x_n \) be the chosen numbers in increasing order. Then:**
   \[
   x_n \geq x_{n-1} + 3 \geq x_{n-2} + 6 \geq \ldots \geq x_1 + 3(n-1)
   \]
   Since these numbers are contained within a set of 2015 consecutive integers, we have:
   \[
   2015 - 1 \geq x_n - x_1 \geq 3(n-1)
   \]
   Simplifying this inequality:
   \[
   2014 \geq 3(n-1) \implies 2014 \geq 3n - 3 \implies 2017 \geq 3n \implies n \leq \frac{2017}{3} \approx 672.33
   \]
   Since \( n \) must be an integer, the maximum value of \( n \) is 672.

7. **Now, we need to show that there exists such a sequence for \( n = 672 \).**

8. **Consider the set \( \{1, 2, \ldots, 2015\} \) and the sequence \( x_k = 3k - 1 \) for \( k \in \{1, 2, \ldots, 672\} \).**

9. **For any two numbers \( x_i \) and \( x_j \) in this sequence:**
   \[
   x_i = 3i - 1 \quad \text{and} \quad x_j = 3j - 1
   \]
   The difference is:
   \[
   x_i - x_j = 3(i - j)
   \]
   which is divisible by 3. The sum is:
   \[
   x_i + x_j = (3i - 1) + (3j - 1) = 3(i + j) - 2
   \]
   which is not divisible by 3.

10. **Thus, the sequence \( x_k = 3k - 1 \) for \( k \in \{1, 2, \ldots, 672\} \) satisfies the condition that no sum of any two selected numbers is divisible by their difference.**

The final answer is \( \boxed{672} \).