An [integer](https://artofproblemsolving.com/wiki/index.php/Integer) between $1000$ and $9999$, inclusive, is called balanced if the sum of its two leftmost [digits](https://artofproblemsolving.com/wiki/index.php/Digit) equals the sum of its two rightmost digits. How many balanced integers are there?
If the common sum of the first two and last two digits is $n$, such that $1 \leq n \leq 9$, there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit).  This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers.  If the common sum of the first two and last two digits is $n$, such that $10 \leq n \leq 18$, there are $19 - n$ choices for both pairs.  This gives $\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285$ balanced numbers.  Thus, there are in total $330 + 285 = \boxed{615}$ balanced numbers.
Both summations may be calculated using the formula for the [sum of consecutive squares](https://artofproblemsolving.com/wiki/index.php/Perfect_square), namely $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.