jmc

If the common sum of the first two and last two digits is $n$, such that $1\le n\le 9$, 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 $10\le n\le 18$, 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, namely ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$.