Let an arbitrary such number be $\overline{a b c d}$. Then, we desire $11 \mid a+b+c+d$ and $11 \mid a-b+c-d$, where the latter comes from the well-known divisibility trick for 11. Sums and differences of multiples of 11 must also be multiples of 11, so this is equivalent to desiring $11 \mid a+c$ and $11 \mid b+d$. As $a \in[1,9]$ and $b, c, d \in[0,9]$, $a+c$ and $b+d$ must be either 0 or 11 (no larger multiple is achievable). There are 8 choices for such $(a, c)$ and 9 choices for such $(b, d)$, so the answer is $8 \cdot 9=72$.