Start with a three-digit positive integer $A$. Obtain $B$ by interchanging the two leftmost digits of $A$. Obtain $C$ by doubling $B$. Obtain $D$ by subtracting $500$ from $C$. Given that $A + B + C + D = 2014$, find $A$.
1. Let \( A = 100a + 10b + c \), where \( a, b, c \) are digits and \( a \neq 0 \) since \( A \) is a three-digit number.
2. By interchanging the two leftmost digits of \( A \), we obtain \( B = 100b + 10a + c \).
3. Doubling \( B \) gives \( C = 2B = 2(100b + 10a + c) = 200b + 20a + 2c \).
4. Subtracting 500 from \( C \) gives \( D = C - 500 = 200b + 20a + 2c - 500 \).
5. Given that \( A + B + C + D = 2014 \), we substitute the expressions for \( A, B, C, \) and \( D \):
   \[
   (100a + 10b + c) + (100b + 10a + c) + (200b + 20a + 2c) + (200b + 20a + 2c - 500) = 2014
   \]
6. Combine like terms:
   \[
   100a + 10b + c + 100b + 10a + c + 200b + 20a + 2c + 200b + 20a + 2c - 500 = 2014
   \]
   \[
   150a + 510b + 6c - 500 = 2014
   \]
7. Add 500 to both sides to simplify:
   \[
   150a + 510b + 6c = 2514
   \]
8. Divide the entire equation by 6 to further simplify:
   \[
   25a + 85b + c = 419
   \]
9. To find \( a, b, \) and \( c \), we start by maximizing \( b \) since it has the largest coefficient. The largest possible value for \( b \) is 4 (since \( b \) is a digit).
   \[
   25a + 85(4) + c = 419
   \]
   \[
   25a + 340 + c = 419
   \]
   \[
   25a + c = 79
   \]
10. Next, we maximize \( a \). The largest possible value for \( a \) is 3 (since \( a \) is a digit and \( a \neq 0 \)).
    \[
    25(3) + c = 79
    \]
    \[
    75 + c = 79
    \]
    \[
    c = 4
    \]
11. Therefore, \( a = 3 \), \( b = 4 \), and \( c = 4 \). Thus, the original three-digit integer \( A \) is:
    \[
    A = 100a + 10b + c = 100(3) + 10(4) + 4 = 344
    \]

The final answer is \(\boxed{344}\).