Develop asymptotic as $N \rightarrow \infty$ of $\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{k} \sum_{d \mid i (2k-i), d > N} 1$

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This calculation arises from expansions of the number of factorable quadratics where the coefficients are constrained by naive height $N$. In the above formula $d$ are the divisors of $i \left({2\, k - i}\right)$ subjected to the additional condition that $d > N$. I do not expect that this sum is reducible. If so that would also be welcomed. My main result that I am looking for is the asymptotic expansion as $N \rightarrow \infty$ to all orders possible. From other parts of this general calculation I have asymptotic terms of ${N}^{2} \log \left({N}\right)^{2}$ that this expansion should also have which will be canceled. This is part of the reason why I need the other terms.

See Asymptotic expansion as $N \rightarrow \infty$ of $\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} k \sum_{e \mid 2k}\frac{\Lambda \left({e}\right)}{e}$ for more details.

The maximum value of $i \left({2\, k - i}\right)$ is $\left\lfloor{N/2}\right\rfloor^{2}$ at $i = {k}_{max} = \left\lfloor{N/2}\right\rfloor$.

This can be reformulated. The source of this problem is the calculation $${S}_{1} \left({N}\right) = \sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{k} \tau \left({i \left({2\, k - i}\right)}\right) - 2 \sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{k} \sum_{d \mid i \left({2\, k-i}\right), d > N}1$$ which comes from the problem of counting the number of reducible quadratics. A slightly modified form which may be easier to expand asymptotically is $${S}_{1} \left({N}\right) = 2 \sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{k} \sum_{d \mid i \left({2\, k-i}\right), d \le N}1 - \sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{k} \tau \left({i \left({2\, k - i}\right)}\right)$$ I have some results for the asymptotic expansion of the $\tau \left({i \left({2\, k - i}\right)}\right)$ (which may or may not be correct since the reference that I am using assumes that for $\tau \left({n \left({n + v}\right)}\right)$ $v$ is a positive integer). I have not be able to expand the first sum in ${S}_{1} \left({N}\right)$ which is a restatement of the posted problem.