The answer is $(-1)^k \binom{n-1}{k}$. Let $f(n,k)=a_{n,k}-b_{n,k}$. The key idea is to pair most odd and even permutations and count number of unpaired ones by getting a recurrence relation. Suppose $\sigma(1)=k$ and for some $l$, $\sigma(l)\ge l$ and $l\le k$. This means swapping $\sigma(l)$ and $\sigma(1)$ creates another permutation with the opposite parity but same $\sum\limits_{j=1}^n |\sigma(j)-j|$. Let $f(\sigma)$ be the smallest integer $l>1$ such that $\sigma(l)\ge l$ and $\sigma(1)\ge l$ (if it doesn't exist $f(\sigma)=-1$). If I swap $\sigma(1), \sigma(l)$, we have $\sigma(j)=j-1$ for all $2\le j\le l-1$, so swapping $\sigma(1), \sigma(l)$ preserves $f(\sigma)$. Furthermore, the operation is an involution, so it is a bijection. (Note swapping any two arbitrarily doesn't make it involution, but this is valid). Hence the number of even permutations with $f(\sigma)=k$ is equal to the number of odd permutations with $f(\sigma)=k$. It follows that $\sigma(j)=j-1$ for all $2\le j\le \sigma(1)$. This means $\sum\limits_{j=1}^{\sigma(1)} |\sigma(j)-j| = 2\sigma(1)$. The parity of the first $\sigma(j)$ elements is $(-1)^{\sigma(j)-1}$ since we have a cycle of length $\sigma(j)$. Therefore, we have the recursion $f(n,k)=\sum\limits_{j=0}^k (-1)^j f(n-1-j,k-j)$. We can check that $f(n,k)=(-1)^k \binom{n-1}{k}$ for small $n,k$ by hand. We can use strong induction: $f(n-1-j,k-j) = (-1)^{k-j} \binom{n-2-j}{k-j}$ then it follows $f(n,k)=\sum\limits_{j=0}^k (-1)^k \binom{n-2-j}{k-j} = (-1)^k \sum\limits_{j=0}^k \binom{n-2-j}{n-2-k} = (-1)^k\binom{n-1}{n-1-k}=(-1)^k\binom{n-1}{k}$.