MathLabs

Problem 1

Let nn be a positive integer. Find the largest nonnegative real number f(n)f(n) such that whenever real numbers a1,a2,…,ana_1,a_2,\ldots,a_n have an integer sum, there is an index ii for which ∣ai−12∣≥f(n)\left|a_i-\frac12\right|\ge f(n).
Step 2 of 5: Assume all deviations are small
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∣ai−12∣<12n (1≤i≤n)⟹∣∑i=1nai−n2∣<12\left|a_i-\frac12\right|<\frac1{2n}\ (1\le i\le n)\Longrightarrow\left|\sum_{i=1}^n a_i-\frac n2\right|<\frac12
Detailed analysis

Now let nn be odd and suppose, for contradiction, that every ∣ai−12∣<12n\left|a_i-\frac12\right|<\frac1{2n}. Summing the strict inequalities gives ∣∑i=1nai−n2∣<12\left|\sum_{i=1}^n a_i-\frac n2\right|<\frac12.