An integer between 1000 and 9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits.  How many balanced integers are there?
To solve the problem, we need to count the number of balanced integers between 1000 and 9999. A balanced integer is defined as an integer where the sum of its two leftmost digits equals the sum of its two rightmost digits.

1. **Define the integer and its digits:**
   Let the integer be represented as \( \overline{abcd} \), where \( a, b, c, \) and \( d \) are its digits. The integer is balanced if:
   \[
   a + b = c + d
   \]

2. **Determine the range of possible sums:**
   Since \( a, b, c, \) and \( d \) are digits (0 through 9), the possible values for \( a + b \) and \( c + d \) range from 0 to 18. However, since \( a \) and \( b \) are the leftmost digits of a four-digit number, \( a \) cannot be 0. Therefore, the minimum value for \( a + b \) is 1, and the maximum value is 18.

3. **Count the number of ways to achieve each sum:**
   We need to count the number of ways to achieve each possible sum \( s \) for both \( a + b \) and \( c + d \). We will denote the number of ways to achieve a sum \( s \) with two digits as \( f(s) \).

   - For \( s = 1 \) to \( s = 9 \):
     \[
     f(s) = s
     \]
     This is because the pairs \((a, b)\) that sum to \( s \) are \((1, s-1), (2, s-2), \ldots, (s-1, 1)\).

   - For \( s = 10 \) to \( s = 18 \):
     \[
     f(s) = 19 - s
     \]
     This is because the pairs \((a, b)\) that sum to \( s \) are \((1, s-1), (2, s-2), \ldots, (9, s-9)\), but we must ensure that both digits are between 0 and 9.

4. **Calculate the total number of balanced integers:**
   For each possible sum \( s \) from 1 to 18, the number of balanced integers is given by the product of the number of ways to achieve \( s \) for the leftmost digits and the number of ways to achieve \( s \) for the rightmost digits:
   \[
   \text{Total balanced integers} = \sum_{s=1}^{18} f(s) \cdot f(s)
   \]

   We can now compute this sum:
   \[
   \sum_{s=1}^{9} s^2 + \sum_{s=10}^{18} (19-s)^2
   \]

   - For \( s = 1 \) to \( s = 9 \):
     \[
     \sum_{s=1}^{9} s^2 = 1^2 + 2^2 + 3^2 + \ldots + 9^2 = \frac{9 \cdot 10 \cdot 19}{6} = 285
     \]

   - For \( s = 10 \) to \( s = 18 \):
     \[
     \sum_{s=10}^{18} (19-s)^2 = 9^2 + 8^2 + 7^2 + \ldots + 1^2 = 285
     \]

   Adding these two sums together:
   \[
   285 + 285 = 570
   \]

The final answer is \(\boxed{570}\).