Asia Pacific Mathematics Olympiad (APMO)

The answer is $$f(n)=\{\begin{array}{ll}0& \text{ if }n\text{ is even }\\ \frac{1}{2n}& \text{ if }n\text{ is odd }\end{array}$$ First, assume that $n$ is even. If $a_i=\frac{1}{2}$ for all $i$, then the sum $a_1+a_2+\cdots +a_n$ is an integer. Since $\left|a_i-\frac{1}{2}\right|=0$ for all $i$, we may conclude $f(n)=0$ for any even $n$.

Now assume that $n$ is odd. Suppose that $\left|a_i-\frac{1}{2}\right|<\frac{1}{2n}$ for all $1\le i\le n$. Then, since ${\sum }_{i=1}^n{a}_i$ is an integer, $$\frac{1}{2}\le \left|\sum _{i=1}^n{a}_i-\frac{n}{2}\right|\le \sum _{i=1}^n\left|a_i-\frac{1}{2}\right|<\frac{1}{2n}\cdot n=\frac{1}{2}$$ a contradiction. Thus $\left|a_i-\frac{1}{2}\right|\ge \frac{1}{2n}$ for some $i$, as required. On the other hand, putting $n=2m+1$ and $a_i=\frac{m}{2m+1}$ for all $i$ gives $\sum a_i=m$, while $$\left|a_i-\frac{1}{2}\right|=\frac{1}{2}-\frac{m}{2m+1}=\frac{1}{2(2m+1)}=\frac{1}{2n}$$ for all $i$. Therefore, $f(n)=\frac{1}{2n}$ is the best possible for any odd $n$.