Answer: 615

Solution:

Group the balanced integers by the common sum $s$ of each digit pair, where $1 \le s \le 18.$ For $s \le 9,$ the leftmost pair (first digit at least $1$) can be formed in $s$ ways and the rightmost pair in $s + 1$ ways. For $s \ge 10,$ both digits of each pair must be at least $s - 9,$ giving $19 - s$ ways for each pair.

The total count is $$\begin{aligned} &\sum_{s=1}^{9} s(s+1) + \sum_{s=10}^{18} (19 - s)^2 \\ &= \sum_{s=1}^{9} (s^2 + s) + \sum_{k=1}^{9} k^2 \\ &= 2 \cdot 285 + 45 = 615. \end{aligned}$$