Vaggelis has a box that contains $2015$ white and $2015$ black balls. In every step, he follows the procedure below:
He choses randomly two balls from the box. If they are both blacks, he paints one white and he keeps it in the box, and throw the other one out of the box. If they are both white, he keeps one in the box and throws the other out. If they are one white and one black, he throws the white out, and keeps the black in the box.
He continues this procedure, until three balls remain in the box. He then looks inside and he sees that there are balls of both colors. How many white balls does he see then, and how many black?
1. **Initial Setup:**
   - Vaggelis starts with 2015 white balls and 2015 black balls.

2. **Procedure Analysis:**
   - **Case 1: Both balls are black**
     - One black ball is painted white and kept in the box.
     - The other black ball is thrown out.
     - The number of black balls decreases by 2.
     - The number of white balls increases by 1.
   - **Case 2: Both balls are white**
     - One white ball is kept in the box.
     - The other white ball is thrown out.
     - The number of white balls decreases by 2.
   - **Case 3: One white and one black ball**
     - The white ball is thrown out.
     - The black ball is kept in the box.
     - The number of white balls decreases by 1.
     - The number of black balls remains the same.

3. **Parity Analysis:**
   - The parity (odd or even nature) of the number of black balls is crucial.
   - Initially, there are 2015 black balls, which is odd.
   - In Case 1, the number of black balls decreases by 2, preserving the odd parity.
   - In Case 2, the number of black balls remains unchanged.
   - In Case 3, the number of black balls remains unchanged.
   - Therefore, the number of black balls will always remain odd.

4. **Final Step:**
   - When only three balls remain, the number of black balls must be odd.
   - The only odd number less than 3 is 1.
   - Hence, there is exactly 1 black ball among the final three balls.

5. **Determining the Number of White Balls:**
   - Since there are three balls in total and one of them is black, the remaining two must be white.

The final answer is $\boxed{2 \text{ white balls and } 1 \text{ black ball}}$