Clearly for $n$ even $f(n)=0$ (as we can choose $a_1=...=a_n=\frac{1}{2}$). Let $n=2k+1$ and make substitution $a_i=\frac{1}{2}+b_i$. We have that $b_1+...+b_{2k+1}=\frac{1}{2}+l$ for some integer $l$. From this we can easy see that $f(2k+1)=\frac{1}{2(2k+1)}$ (suppose that for all $i$ we have $|b_i| < \frac{1}{2(2k+1)}$ to arrive at contradiction)

First notice that : $f(n) = \min_{a\in \mathbb{R}^n / \sum_{i=1}^{n}a_i \in \mathbb{Z}} (\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i|)$. If n is even, we just need to choose $a_1=a_2=\cdots=a_n=\frac{1}{2}$ to see that f(n)=0. If n is odd, write n=2k+1. If $\frac{k}{2k+1} < a_1, a_2, \cdots, a_n < \frac{k+1}{2k+1}$, summing up we get : $k < \sum_{i=1}^{n}a_i < k+1$ which contradicts $\sum_{i=1}^{n}a_i \in \mathbb{Z}$. Hence there exists an $a_i$ such that $a_i \not \in (\frac{k}{2k+1},\frac{k+1}{2k+1})$ and hence $\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i| \geq |\frac{1}{2}-\frac{k}{2k+1}| = |\frac{1}{2n}|$. Notice also that $a_1=a_2=\cdots=a_{2k+1}=\frac{k}{2k+1}$ leads to $\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i| = |\frac{1}{2n}|$. Hence : $f(n) = \min_{a\in \mathbb{R}^n / \sum_{i=1}^{n}a_i \in \mathbb{Z}} (\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i|) = |\frac{1}{2n}|$.

first , we shall show that , for even $n$ , $f(n)=0$ . pf-- clearly $f(n) \ge 0$ . for any even $n$ , consider the case when all $a_i$'s are $1/2$ , then $\sum {a_i}$ is an integer . then $\mid a_i-1/2 \mid =0$ for all $i=1,2,...,n$ . hence $0 \ge f(n)$ . so , $f(n)=0$ , which indeed satisfies the conditions. next we show that , for odd $n$ , $f(n)=\frac{1}{2n}$ consider any odd $n$ . then $\sum{\mid a_i-1/2 \mid} \ge \mid \sum{a_i}-\frac{n}{2} \mid \ge 1/2$ . so , there exists some $j \in$ {1,2,...,n} , such that , $\mid a_{j}-1/2 \mid \ge 1/2n$ . hence $f(n) \ge 1/2n$ . now consider the case when all $a_{i}$'s are equal to $1/2+1/2n$ . then $a_1+a_2+....+a_n$ is an integer and $\mid a_{i}-1/2 \mid =1/2n$ for all $i \in $ {1,2,..,n}. hence , $f(n)=\frac{1}{2n}$ for all odd $n$

Dijkschneier wrote: First notice that : $f(n) = \min_{a\in \mathbb{R}^n / \sum_{i=1}^{n}a_i \in \mathbb{Z}} (\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i|)$. If n is even, we just need to choose $a_1=a_2=\cdots=a_n=\frac{1}{2}$ to see that f(n)=0. If n is odd, write n=2k+1. If $\frac{k}{2k+1} < a_1, a_2, \cdots, a_n < \frac{k+1}{2k+1}$, summing up we get : $k < \sum_{i=1}^{n}a_i < k+1$ which contradicts $\sum_{i=1}^{n}a_i \in \mathbb{Z}$. Hence there exists an $a_i$ such that $a_i \not \in (\frac{k}{2k+1},\frac{k+1}{2k+1})$ and hence $\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i| \geq |\frac{1}{2}-\frac{k}{2k+1}| = |\frac{1}{2n}|$. Notice also that $a_1=a_2=\cdots=a_{2k+1}=\frac{k}{2k+1}$ leads to $\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i| = |\frac{1}{2n}|$. Hence : $f(n) = \min_{a\in \mathbb{R}^n / \sum_{i=1}^{n}a_i \in \mathbb{Z}} (\max_{i\in \{1,2,\cdots,n\}} |\frac{1}{2}-a_i|) = |\frac{1}{2n}|$. The first part of your proof is pretty good, but I'm not entirely sure that you can state the last two claims (the max min stuff through equality) without providing a basic smoothing argument (cause there are two configs that sum to k+1 and k resp., and it doesn't seem like you address the case of $a_i = \frac{k+1}{2k+1}$ in the last part.) I am thinking somewhere along the lines of this: 1) Prove that $\sum_{i=1}^{n}a_i$ = k+1 or k. (Assume k+1 config, if sum is greater than k+1, then one $a_i$ must be greater than $\frac{k+1}{2k+1}$, contradiction. Repeat a similar argument for k config) 2) For k+1, and k config, since n is fixed and sum is fixed (k or k+1), one element greater than the equality case implies another is less than the equality case. Use this to get contradiction.

I don't quite understand the question though. Isn't it if we plot a_1=1/2 and we put the rest arbritrary then we get f(n)=0?

Oh ok i get it now sorry fr interupting