Consider two adjacent numbers \( a \geq b \), since they differ at least in \( 2 \) times, thus \( a \geq 2b \). Therefore, \( 3(a - b) = (a + b) + 2(a - 2b) \geq (a + b) \). This indicates that the sum of differences is at least one-third of the sum of sums, and the sum of sums of each pair of adjacent numbers is twice the total sum of all numbers, which is \( 2 \). This implies that the total sum of differences is at least \( \boxed{\frac{2}{3}} \).

lbh_qys wrote: Consider two adjacent numbers \( a \geq b \), since they differ at least in \( 2 \) times, thus \( a \geq 2b \). Therefore, \( 3(a - b) = (a + b) + 2(a - 2b) \geq (a + b) \). This indicates that the sum of differences is at least one-third of the sum of sums, and the sum of sums of each pair of adjacent numbers is twice the total sum of all numbers, which is \( 2 \). This implies that the total sum of differences is at least \( \boxed{\frac{2}{3}} \). You have to find a construction. Anyway, $$\frac{1}{3036}, \frac{2}{3036}, \frac{1}{3036}, \frac{2}{3036}, \ldots$$works.