This expression comes from the asymptotic expansion of
$$\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \sum_{i=1}^{\left\lfloor{\sqrt{{N}^{2}+{k}^{2}}}\right\rfloor - k} \tau \left({i \left({2\, k + i}\right)}\right)$$
From Adrian W. Dudek, "Note on the number of divisors of reducible quadratic polynomials", arXiv:1806.01404v1 [math.NT] 4 Jun 2018. Here $\tau \left({x}\right)$ us the number of divisors of $x$ and $\left\{{x}\right\}$ is the fractional part of $x$.
Now assuming that $\sqrt{{N}^{2} + {k}^{2}}$ is distributed uniformly we can write assuming that $\left\{{x}\right\} = 1/2$ as $N \rightarrow \infty$
$$S \left({N}\right) = \sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \left\{{\sqrt{{N}^{2} + {k}^{2}}}\right\} \sim \frac{1}{4} N$$
The following numerical tests are observed
$$\begin{array}{ccc} N & S \left({N}\right) & N/4 - S \left({N}\right) \\ {10}^{4} & 2,470.1789390926911658 & 29.8210609073088342 \\ {10}^{5} & 24,907.909459652367788 & 92.090540347632212 \\ {10}^{6} & 249,660.30707854707527 & 339.69292145292473 \\ {10}^{7} & 2,498,926.5245972764825 & 1,073.4754027235175 \\ {10}^{8} & 24,996,910.832762803688 & 3,089.167237196312 \end{array}$$
which indicates that the actual expansion is
$$S \left({N}\right) \sim \frac{1}{4} N - \frac{1}{3} \sqrt{N} + \cdots$$
I am looking for a derivation of this result assuming that my approximation it is correct.
Applying the definition of the floor function we get $$\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \left\{{\sqrt{{N}^{2} + {k}^{2}}}\right\} = \frac{1}{2}\, \left\lfloor{\frac{N}{2}}\right\rfloor -\frac{1}{2} \sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} \left[{\sqrt{{N}^{2}+{k}^{2}} \in \mathbb{Z}}\right] - \frac{1}{\pi} \sum_{k = 1,\sqrt{{N}^{2}+{k}^{2}} \not\in \mathbb{Z}}^{\left\lfloor{N/2}\right\rfloor} arctan \left({\cot \left({\pi \sqrt{{N}^{2} + {k}^{2}}}\right)}\right)$$ where $\left[{...}\right]$ are Iverson brackets. Due to the numerous jump disconnectedness we can not apply the Euler-Maclaurin formula to the trigonometric sum.
Plot of the $\sum_{k=1}^{\left\lfloor{N/2}\right\rfloor} arctan \left({\cot \left({\sqrt{{N}^{2}+{k}^{2}}}\right)}\right)$function showing the discontinuities with estimated envelop bands $\pm \sqrt{N}/3$
I arrived at this formula by using Mathematica Version 12 applied to the definition of the fractional part to obtain the closed form $$\left\{{\sqrt{{N}^{2} + {k}^{2}}}\right\} = \frac{1}{2} - \frac{1}{\pi} \sum_{n = 1}^{\infty} \frac{1}{n} \sin \left({2\, \pi\, n \sqrt{{N}^{2} + {k}^{2}}}\right) = \frac{1}{2} - \frac{i}{2\, \pi} \left({\log \left({1 - {e}^{2\, \pi\, i \sqrt{{N}^{2} + {k}^{2}}}}\right) - \log \left({1 - {e}^{- 2\, \pi\, i \sqrt{{N}^{2} + {k}^{2}}}}\right)}\right) = \frac{1}{2} - \frac{1}{\pi} arctan \left({\cot\left({\pi \sqrt{{N}^{2} + {k}^{2}}}\right)}\right)$$
where $\sqrt{{N}^{2}+{k}^{2}} \not\in \mathbb{Z}$ after simplification to real variables.
This is where the problem is derived: $$\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \left\lfloor{\sqrt{{N}^{2} + {k}^{2}}}\right\rfloor = \sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sqrt{{N}^{2} + {k}^{2}}- \sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \left\{{\sqrt{{N}^{2} + {k}^{2}}}\right\}$$
I have used Euler-Maclaurin formula on the second sum of the above equation to get
$$\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \sqrt{{N}^{2} + {k}^{2}} \sim \frac{1}{8} \left({\sqrt{5} + 4\, arccsch \left({2}\right)}\right) {N}^{2} + \frac{1}{4} \left({\sqrt{5} - 2}\right) N + \frac{1}{12 \sqrt{5}} + \left({{N}^{- 2}}\right)$$
Which leaves the fractional part sum. If on the other hand a better solution can be derived from the first sum $\sum_{k = 1}^{\left\lfloor{N/2}\right\rfloor} \left\lfloor{\sqrt{{N}^{2} + {k}^{2}}}\right\rfloor$ that would do just as well. The first sums looks like it may be a version of Gauss Circle Problem.