It seems possible. The examples which (to me) seem extreme with $N=3,4,5,6$ are ok and the gap is growing (starting at $N=4$) so perhaps estimation would suffice for the last two cases or perhaps they would not be hard to find exactly.

• For $N=3$ one has $2\sqrt(3)-(2+\sqrt{2}) \approx 0.04988805$
• For $N=4$ one has $2\sqrt(4)-4=0$
• For $N=5$ one has $2\sqrt(5)-(3+2\sqrt{1/2} \approx 0.05792239$ (corners and center)
• For $N=6$ one has $2\sqrt{6}-4.5 \approx 0.398979$

The last is more of a guess. It is the result of choosing the four corners and two points on the middle line so that the two paths shown are equal in length. Unexpectedly (to me) that puts the two internal points at $(\frac{4 \pm 1}{8},\frac{1}{2}).$ I suppose that the scaled $3-4-5$ right triangle is not so surprising.

It seems possible. The examples which (to me) seem extreme with $N=3,4,5,6$ are ok and the gap is growing (starting at $N=4$) so perhaps estimation would suffice for the last two cases or perhaps they would not be hard to find exactly.

• For $N=3$ one has $2\sqrt3-(2+\sqrt{2}) \approx 0.04988805$
• For $N=4$ one has $2\sqrt4-4=0$
• For $N=5$ one has $2\sqrt5-(3+2\sqrt{1/2} \approx 0.05792239$ (corners and center)
• For $N=6$ one has $2\sqrt{6}-4.5 \approx 0.398979$

The last is more of a guess. It is the result of choosing the four corners and two points on the middle line so that the two paths shown are equal in length. Unexpectedly (to me) that puts the two internal points at $(\frac{4 \pm 1}{8},\frac{1}{2}).$ I suppose that the scaled $3-4-5$ right triangle is not so surprising.