We show that $2$ works, and then we prove that it is minimum. Define the sequence as follows: Fill the first position with any integer. Then, fill the two rightmost unoccupied cells with $\frac{1}{2}$. Then fill the two leftmost unoccupied cells with $\frac{1}{2}$. Alternate between the left most and right most extremes, till only $1$ cells remains. Fill that cell with any integer. For example, consider $n = 8$ The sequence would be constructed as follows ($a$ represents any integer): $a, , , , , , , $ $a, , , , , , \frac{1}{2}, \frac{1}{2}$ $a, \frac{1}{2}, \frac{1}{2}, , , , \frac{1}{2}, \frac{1}{2}$ $a, \frac{1}{2}, \frac{1}{2}, , \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ $a, \frac{1}{2}, \frac{1}{2}, a, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ It is easy to see that this works. Now, to prove that $2$ is the minimum: ($x_i$ represents a real number, not an integer and $a_i$ represents an integer) Obviously, the first/last cell has to be an integer $(= a_0)$. WLOG, the first cell is an integer. This forces us to fill the last two cells with $x_1$ and $a_1-x_1$, because the first $2$ cells cannot be an integer unless the second cell itself is an integer. This forces the 2nd and 3rd cell to be filled with $x_2$ and $a_2 - x_2$, because the sum of the last $3$ cells cannot be an integer unless the third from last cell is an integer. This process continues, till we are only left with $1$ cell. The remaining $(n-1)$ cells sum to be an integer, as the sum is: $a_0 + \sum_{j=1}^{i} a_j - x_j + \sum_{j=1}^{i}x_j = \sum_{j=0}^{i}a_j$ which is an integer. Now this last cell has to be filled with an integer, because the sum of all $n$ cells must be an integer. Hence $2$ is the minimum