Which number we need to substract from numerator and add to denominator of $\frac{\overline{28a3}}{7276}$ such that we get fraction equal to $\frac{2}{7}$
1. We start with the given equation:
   \[
   \frac{\overline{28a3} - x}{7276 + x} = \frac{2}{7}
   \]
   Here, $\overline{28a3}$ represents a four-digit number where $a$ is a digit.

2. Cross-multiplying to eliminate the fraction, we get:
   \[
   7(\overline{28a3} - x) = 2(7276 + x)
   \]

3. Expanding both sides:
   \[
   7 \cdot \overline{28a3} - 7x = 2 \cdot 7276 + 2x
   \]

4. Simplifying the right-hand side:
   \[
   7 \cdot \overline{28a3} - 7x = 14552 + 2x
   \]

5. Rearranging the equation to isolate $x$:
   \[
   7 \cdot \overline{28a3} - 14552 = 9x
   \]

6. Solving for $x$:
   \[
   x = \frac{7 \cdot \overline{28a3} - 14552}{9}
   \]

7. To ensure $x$ is an integer, $7 \cdot \overline{28a3} - 14552$ must be divisible by 9. We need to find the value of $a$ that satisfies this condition.

8. Let's substitute $\overline{28a3}$ with $2803 + 100a$:
   \[
   7 \cdot (2803 + 100a) - 14552 = 9x
   \]

9. Simplifying the left-hand side:
   \[
   19621 + 700a - 14552 = 9x
   \]
   \[
   5069 + 700a = 9x
   \]

10. For $9x$ to be an integer, $5069 + 700a$ must be divisible by 9. We need to find the value of $a$ such that this condition holds.

11. Checking for $a = 1$:
    \[
    5069 + 700 \cdot 1 = 5769
    \]
    \[
    5769 \div 9 = 641
    \]
    Since 641 is an integer, $a = 1$ is a valid solution.

12. Substituting $a = 1$ back into the equation:
    \[
    x = \frac{5069 + 700 \cdot 1}{9} = \frac{5769}{9} = 641
    \]

Therefore, the value of $x$ is $\boxed{641}$.