1. Let the distances of the two parallel chords from the center be $d_1$ and $d_2$.

   For a circle of radius $R=2$, if a chord subtends angle $\theta$ at the center, then its distance from the center is $d=R\cos\frac{\theta}{2}.$

2. Here the chords subtend angles $\frac{\pi}{k} \quad \text{and} \quad \frac{2\pi}{k}.$

   So their distances from the center are $d_1=2\cos\left(\frac{1}{2}\cdot \frac{\pi}{k}\right)=2\cos\frac{\pi}{2k},$ $d_2=2\cos\left(\frac{1}{2}\cdot \frac{2\pi}{k}\right)=2\cos\frac{\pi}{k}.$

3. Since the chords are parallel and the distance between them is $\sqrt{3}+1$, there are two possibilities:

   • chords on the same side of the center: $|d_1-d_2|=\sqrt{3}+1$
   • chords on opposite sides of the center: $d_1+d_2=\sqrt{3}+1$

   Now, $d_1=2\cos\frac{\pi}{2k}, \quad d_2=2\cos\frac{\pi}{k}.$

   Using $\cos\frac{\pi}{2k} \ge \cos\frac{\pi}{k}$, the difference is at most $2$, while the sum can be as large as $4$.

4. Check the same-side case: $2\cos\frac{\pi}{2k}-2\cos\frac{\pi}{k}=\sqrt{3}+1.$

   Try the natural standard angles. If $\frac{\pi}{2k}=\frac{\pi}{6}, \quad \frac{\pi}{k}=\frac{\pi}{3},$ then $d_1=2\cos\frac{\pi}{6}=2\cdot \frac{\sqrt3}{2}=\sqrt3,$ $d_2=2\cos\frac{\pi}{3}=2\cdot \frac12=1.$ Hence $d_1+d_2=\sqrt3+1,$ not the difference.

   So the chords must lie on opposite sides of the center.

5. Therefore, $2\cos\frac{\pi}{2k}+2\cos\frac{\pi}{k}=\sqrt3+1.$

   From the observation above, $\cos\frac{\pi}{2k}=\frac{\sqrt3}{2}, \qquad \cos\frac{\pi}{k}=\frac12$ gives exactly $2\cdot \frac{\sqrt3}{2}+2\cdot \frac12=\sqrt3+1.$

   This implies $\frac{\pi}{2k}=\frac{\pi}{6}, \qquad \frac{\pi}{k}=\frac{\pi}{3},$ and both give $k=3.$

6. Hence, $[k]=[3]=3.$

Therefore the required integer is $3$.