We prove the case $D = \frac{5\sqrt{2}}{4}$. Let us arrange $\{1, 2, \dots, N\}$ into a triangle as in the figure. For any point $X$ in the triangle, let $r_X$ be the row number counted from the vertex $A$. Let $Y$ and $Z$ be the points corresponding to $P$ after rotating the triangle clockwise and counterclockwise, respectively. Since $ABC$ is an equilateral triangle, we have
$$r_X + r_Y + r_Z = 2m + 1 = \sqrt{8N + 1}.$$
Furthermore, let us consider $n_X$ to be the number corresponding to $X$ in the triangle. Since the top $r$ rows of the triangle contain a total of $1 + 2 + \dots + r = \frac{r(r+1)}{2}$ points, we have
$$\frac{r_X(r_X - 1)}{2} + 1 \le n_X \le \frac{r_X(r_X + 1)}{2},$$
and hence
$$\left( r_X - \frac{1}{2} \right)^2 < n_X < \left( r_X + \frac{1}{2} \right)^2 .$$
Therefore we have
$$\left| \sqrt{2n_X} + \sqrt{2n_Y} + \sqrt{2n_Z} - (r_X + r_Y + r_Z) \right| < \frac{3}{2},$$
that is,
$$\left| \sqrt{n_X} + \sqrt{n_Y} + \sqrt{n_Z} - 2\sqrt{N + \frac{1}{8}} \right| < \frac{3}{2} \times \frac{1}{\sqrt{2}} < \frac{3\sqrt{2}}{4}.$$
Also,
$$2\sqrt{N + \frac{1}{8}} - 2\sqrt{N} < 2\sqrt{\frac{1}{8}} = \frac{2\sqrt{2}}{4},$$
so we have
$$\left| \sqrt{n_X} + \sqrt{n_Y} + \sqrt{n_Z} - 2\sqrt{N} \right| < \left( \frac{3}{4} + \frac{2}{4} \right) \sqrt{2} = \frac{5\sqrt{2}}{4}.$$

Therefore, letting $n_X$ range from 1 to $N$, $(n_X, n_Y, n_Z)$ satisfies what is required in the problem.