This restricted divisor problem is derived from counting the number of reducible quadratics where we have for one of the four main terms
$$\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i = 1}^{k} \tau \left({i \left({2\, k - i}\right)}\right) - 2 \sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i = 1}^{k} \sum_{\substack{d \mid i \left({2\, k - i}\right), \\ d > N}} 1 = \sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i = 1}^{k} \sum_{\substack{{d}_{1}\, {d}_{2} = i \left({2k - i}\right), \\ {d}_{1} \le N, {d}_{2} \le N}} 1 =\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i = 1}^{k} \sum_{{d}_{1} = 1}^{N} \sum_{{d}_{2} = 1}^{N} \left[{{d}_{1}\, {d}_{2} = i \left({2\, k - i}\right)}\right]$$
where $\tau \left({n}\right)$ is the number of divisors of $n$, $N \ge 1$, $\left[{...}\right]$ is the Iverson bracket, and $d$, ${d}_{1}$, and ${d}_{2}$ are divisors of $i \left({2\, k - i}\right)$.
Can this last sum be simplified and by what is the asymptotic expansion as $N \rightarrow \infty$.
In J. L. Truelsen "Divisor problems and the pair correlation for the fractional parts of ${n}^{2}\, \alpha$", Intern. Math Research Notices, 2010, Issue 16, p 3144-3183, introduces the function $${\tau}_{M, N} \left({m}\right) = \# \left\{{\left({a, b}\right) \mid 1 \le a \le M, 1 \le b \le N: a\, b = m}\right\}$$
and when $M = N$, the function ${\tau}_{N}^{*} \left({m}\right) = {\tau}_{N, N} \left({m}\right)$. In terms of Truelsen's function we can write $$\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i = 1}^{k} \sum_{\substack{{d}_{1}\, {d}_{2} = i \left({2k - i}\right), \\ {d}_{1} \le N, {d}_{2} \le N}} 1 = \sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i = 1}^{k} {\tau}_{N}^{*} \left({i \left({2\, k - i}\right)}\right)$$
So expanding Truelsen's divisor function may be the solution. However, this work is only 10 years old and there does not appear to be much published on this function. And the usage in his paper is for a limited case.