Alternating sum of hook lengths: Part II This is a follow up on my earlier MO post.
Given $\lambda$ an integer partition of $n$, let $h_{ij}(\lambda)$ denote the hook length of cell $(i,j)$ in the Young diagram of $\lambda$. Let
$$f_n=\sum_{\lambda\vdash n}\sum_{(i,j)\in\lambda}(-1)^{i+j}\cdot h_{ij}(\lambda).$$
This sequence $f_n$ shows up on OEIS A066897 as "total number of odd parts in all partitions of $n$. So, I ask:

QUESTION. What is a bijection between this hook length statistic and odd parts of partitions?

 A: There are a couple of things at play here. One is the fact that for each partition $\lambda$ we have
$$\sum_{(i,j)\in\lambda}(-1)^{i+j}\cdot h_{ij}(\lambda)=\frac{(\lambda_1-\lambda_2+\lambda_3+\cdots)+(\lambda'_1-\lambda'_2+\lambda'_3+\cdots)}{2}.$$
This is easy to see from the description I mentioned in the answer to your previous question: this counts the number of cells with both coordinates even minus the cells with both coordinates odd. Now if you pair each partition with its conjugate, you see that summing this statistic over all partitions of $n$ is equivalent to simply summing the statistic $\lambda_1-\lambda_2+\lambda_3+\cdots$. This is as bijective as this argument will get because these two statistics aren't equally distributed among partitions of $n$.
Now it turns out that this last statistic $\lambda_1-\lambda_2+\lambda_3+\cdots$ is equally distributed to the statistic "number of odd parts", and this can be proven bijectively. In fact this is essentially exercise 3.4.5 in Igor Pak's "Partition Bijections: A Survey" (you would have to combine it with Franklin's bijection as described in 3.3.1 to get the full result).
