Solution:
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 $8 \cdot 9 = 72$.