Let $(1, 1), (2N, 2N)$ be the bottom-left and the top-right corner, respectively, and $f(i)$ be the $i$-th cell on the route. (a) Since every domino can contribute at most $1$ longitudinal move, the maximum possible number of longitudinal moves $\leq2N^2$. Construction of $2N^2$: $f(i)=(\lceil\frac i{2N}\rceil, n+\frac12-(-1)^{\lceil\frac i{2N}\rceil}(i-n-\frac12))$, dominos: $\{\{(i, 2j-1), (i, 2j)\}|1\leq i\leq 2N,\ 1\leq j\leq N\}$ $\therefore$ the answer is $2N^2$. (b) For $N=1$, the answer is trivially $1$. For $N\geq2$: There are at least two corners $f(a), f(b)$ that are neither $f(1)$ nor $f(4N^2)$. Since one of $f(a\pm1)$ and one of $f(b\pm1)$ is on the same domino as $f(a), f(b)$, respectively. $\therefore$ the minimum possible number of longitudinal moves $\geq2$ since for $N\geq2$, no domino can cover two corners ($f(a), f(b)$) simultanously. Construction of $2$: $f(i)=(\lceil\frac i{2N}\rceil, n+\frac12-(-1)^{\lceil\frac i{2N}\rceil}(i-n-\frac12))$, dominos: $\{\{(2i-1, j), (2i, j)\}|1\leq i\leq N,\ 1\leq j\leq2N-2\}\cup\{\{(2i, j), (2i+1, j)\}|1\leq i\leq N-1,\ 2N-1\leq j\leq2N\}\cup\{\{(1, 2N-1), (1, 2N)\}, \{(2N, 2N-1), (2N, 2N)\}\}$ $\therefore$ the answer is $2$ if $N\geq2$.