Eric has $2$ boxes of apples, with the first box containing red and yellow apples and the second box containing green apples. Eric observes that the red apples make up $\tfrac{1}{2}$ of the apples in the first box. He then moves all of the red apples to the second box, and observes that the red apples now make up $\tfrac{1}{3}$ of the apples in the second box. Suppose that Eric has $28$ apples in total. How many red apples does Eric have?
1. Let the number of red apples be \( r \), yellow apples be \( y \), and green apples be \( g \).
2. Since the red apples make up \(\frac{1}{2}\) of the apples in the first box, and the only other color in the first box is yellow, we have:
   \[
   r = y
   \]
3. After moving all the red apples to the second box, Eric observes that the red apples now make up \(\frac{1}{3}\) of the apples in the second box. This implies that the remaining \(\frac{2}{3}\) of the apples in the second box are green. Therefore, we have:
   \[
   r = \frac{1}{3}(r + g) \implies 3r = r + g \implies g = 2r
   \]
4. We are given that Eric has 28 apples in total. Therefore, we can write the equation:
   \[
   r + y + g = 28
   \]
5. Substituting \( y = r \) and \( g = 2r \) into the equation, we get:
   \[
   r + r + 2r = 28 \implies 4r = 28 \implies r = 7
   \]

The final answer is \( \boxed{7} \)