Solution:

Let $k = |\mathcal{A}|,$ so $|\mathcal{B}| = 12 - k$ with $1 \le k \le 11.$ Since every element lies in exactly one set, $k \notin \mathcal{A}$ means $k \in \mathcal{B},$ and $12 - k \notin \mathcal{B}$ means $12 - k \in \mathcal{A}.$ If $k = 6,$ then $6$ would have to belong to both sets, which is impossible, so $k \ne 6.$

For each other $k,$ the elements $k$ and $12 - k$ are already placed, and the remaining $k - 1$ elements of $\mathcal{A}$ can be chosen from the other $10$ numbers in $\binom{10}{k-1}$ ways, with $\mathcal{B}$ taking the rest. Hence $$N = \sum_{k=1}^{11} \binom{10}{k-1} - \binom{10}{5} = 2^{10} - 252 = 772.$$