Problem 1

Let

n

be a positive integer. Find the largest nonnegative real number

f(n)

such that whenever real numbers

a_1,a_2,\ldots,a_n

have an integer sum, there is an index

i

for which

\left|a_i-\frac12\right|\ge f(n)

Step 4 of 5: Attain the odd-case bound

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n=2m+1,\qquad a_i=\frac{m}{2m+1}\quad(1\le i\le n)

Detailed analysis

Write $n=2m+1$ and set every $a_i=\frac{m}{2m+1}$ . Then the sum is the integer $m$ , and each distance is $\left|a_i-\frac12\right|=\frac1{2(2m+1)}=\frac1{2n}$ . Thus the lower bound is sharp.