Lemma 1: Assume that we have $a_1\geq a_2\geq \ldots a_n\geq 0, 0 \leq b_1\leq b_2\leq \ldots b_n$ with $a_1+a_2+\ldots+a_k\leq b_k$ for $1\leq k \leq n$. Then $a_1^2+\ldots+a_n^2\leq b_1^2+(b_2-b_1)^2+\ldots+(b_n-b_{n-1})^2$. Proof: if suffices to apply Karamata for the convex and increasing function $x^2$. Lemma 2: The sum of all numbers less than $n$ and of the same parity with $n-1$ is $\frac{n(n^2-1)}6$. This is proven by induction. Let $S=\sum_{1\leq i<j\leq n} (b_i-b_j)^2=n(b_1^2+\ldots+b_n^2)-(b_1+\ldots+b_n)^2$ Assume that $b_1,b_2,\ldots,b_n$ realize the minimum. It's clear that we can replace all $b_i$ by $b_i-t$ invariating $S$, so we can assume that $b_1+\ldots+b_n=0$. Now let $k$ of the numbers $b_i$ be positive and $l=n-k$ of them be negative, with $k \geq \frac n2$. Assume that we subtract $1$ from $p$ of the positive numbers. $S$ then will change by $np-p^2-2ns$ where $s$ is the sum of the $p$ numbers. Since $S$ is minimal, this value is non-negative. Therefore we deduce that the sum of any $p$ positive numbers is at most $\frac{p(n-p)}{2}$. By letting $p=k$ we deduce that the sum of the all positve numbers is at most $\frac{k(n-k)}2$. So the sum of any $p$ positive numbers is at most ${min \{\frac{k(n-k)}2, \frac{p(n-p)}2}\}$. Now applying Lemma 1 for these numbers we get that the sum of squares of the positive numbers is at most: $\sum_{i=1}^{n-k} (\frac{i(n-i)}{2n}-\frac{(i-1)(n-i+1)}2n)^2=\sum_{i=1}^{n-k} \frac{(n-2i+1)^2}{4n^2} \leq \frac 1{4n^2}\sum i^2)$, where the sum is taken over all $i$ less than $n$ of the same parity with $n-1$. Applying Lemma 2, this sum is at most $\frac{n(n^2-1)}{24n^2}=\frac{n(n^2-1)}{24n}$. Reasoning analogously for the negative $b_i$ (just now add $1$ instead of substracting), we deduce that the sum of squares of the negative numbers is also at most $\frac{n(n^2-1)}{24n}$. Hence $S=n(b_1^2+\ldots+b_n^2)\leq n(\frac{n(n^2-1)}{24n}+\frac{n(n^2-1)}{24n})=\frac{n^2-1}{12}$. QED Happy Easter!

why can you $b_1+b_2+\cdots+b_n=0$ that means $a_1+a_2+\cdots+a_n$ is integer. :

shobber wrote: Given $ n$ real numbers $ a_1$, $ a_2$ $ \ldots$ $ a_n$. ($ n\geq 1$). Prove that there exists real numbers $ b_1$, $ b_2$ $ \ldots$ $ b_n$ satisfying: (a) For any $ 1 \leq i \leq n$, $ a_i - b_i$ is a positive integer. (b)$ \sum_{1 \leq i < j \leq n} (b_i - b_j)^2 \leq \frac {n^2 - 1}{12}$ It is obvious that if there exists reals $ b_i$ such that $ a_i - b_i$ is an integer, then there exists reals $ b_i$ such that $ a_i - b_i$ is a positive integer with the same sum of squares of difference - thus not changing (b). So replace (a) with (c) For any $ 1 \leq i \leq n$, $ a_i - b_i$ is an integer. Let a sequence $ \underline{c} = (c_1,c_2,..,c_n)$ be "valid" if there exists $ b_1,b_2,...,b_n$ such that $ c_i = nb_i - \sum_{i = 1}^n b_i$, and $ b_i - a_i$ is an integer. Define the transformation $ \Delta_i$ as: $ \underline{c} = (c_1,c_2,c_3, \cdots, c_n)$ $ \to (c_1 - 1,c_2 - 2, \cdots, c_{i - 1} - 1, c_i + n - 1, c_{i + 1} - 1, \cdots, c_n - 1) = \underline{c'}$ Then $ \underline{c'}$ is obviously a valid sequence if $ \underline{c}$ is. Let $ \underline{d} = (d_1,..,d_n), d_i = na_i - \sum_{j = 1}^n a_j$. Define $ f(\underline{x}) = \sum_{i = 1}^n x_i^2$, and let $ f$ be defined on all sequences $ \underline{x}$ which can be obtained from $ \underline{d}$ with a finite number of $ \Delta_i$ operations. First I'll show that $ f(\underline{x})$ has a global minimum. Let $ \underline{y} = (y_1, \cdots, y_n)$ be an permutation of $ \underline{x}$ such that $ y_1 \le y_2 \le \cdots \le y_n$. If $ y_n - y_1 > n^2$ there must exists an $ k$ such that $ y_{k + 1} - y_k > n$. Then consider $ \underline{x'}$, which is $ \underline{x}$ after the operations: $ \Delta_1\Delta_2\cdots\Delta_k$. As $ \sum_{i = 1}^n x_i = 0$ we find that: $ f(\underline{x}) = \sum_{i = 1}^n x_i^2 = \frac {1}{n} \sum_{1 \le i < j \le n} (x_i - x_j)^2$. From that and the fact that $ \underline{x'}$ is a "contraction" of $ \underline{x}$ it is easy to see that $ f(\underline{x'}) < f(\underline{x})$. Furtermore the difference between the biggest and smallest element in the two sequences $ \underline{x}, \underline{x'}$ differ by exactly $ n$. So we can form a sequence where we can get from $ \underline{x_i}$ to $ \underline{x_{i + 1}}$ with the operations $ \Delta_1 \cdots \Delta_{k_i}$ for some $ k_i$, such that the difference between the greatest and smallest element in $ \underline{x_n}$ is at most $ n^2$: $ \underline{x} = \underline{x_0} \to \underline{x_1} \to \cdots \to \underline{x_n} = \underline{x''}$ such that the difference between the biggest and the smallest element of $ \underline{x''}$. Since $ \underline{x''} = (x''_1, \cdots, x''_n)$. Since $ \sum_{i = 1}^n x''_i = 0$, we see $ |x''_i| \le n^2$. And since the fractional part of $ x''_i$ is given by $ a_i$, $ \underline{x''}$ belongs to a finite set. There is clearly a global minimum in this finite set, and therefore $ f(\underline{x})$ has a global minimum. Let $ \underline{m}$ be the global minimum of $ f$. I'll prove that $ f(\underline{m}) \le \frac {n(n^2 - 1)}{12}$. $ \underline{m} = (m_1,m_2,\cdots,m_n)$ Then $ \underline{m'} = \Delta_1\Delta_2\cdots\Delta_k\underline{m} =$ $ (m_1 + (n - k),m_2 + (n - k), \cdots, m_k + (n - k), m_{k + 1} - k, m_{k + 2} - k, \cdots, m_n - k)$ Then $ f(\underline{m'}) - f(\underline{m}) =$ $ \sum_{i = 1}^k 2(n - k)m_i + (n - k)^2 + \sum_{i = k + 1}^n - 2km_i + k^2 =$ $ 2n\sum_{i = 1}^k m_i + k(n - k)n \ge 0 \iff$ $ \sum_{i = 1}^k m_i \ge - \frac {k(n - k)}{2} \iff$ $ \sum_{i = k + 1}^n m_i \le \frac {k(n - k)}{2} = \frac {(n - k)(n - (n - k))}{2}$. Since we could have changed $ \Delta_1\cdots\Delta_k$ by $ \Delta_{i_1}\cdots\Delta_{i_k}$ for any abritary different indices $ i_1, \cdots, i_k$, we find: $ \left | \sum_{j = 1}^k m_{i_j} \right | \le \frac {k(n - k)}{2}$ Consider the sequence $ \underline{M} = (M_1, \cdots, M_n), M_i = \frac {n + 1}{2} - \frac {i}{n}$. Then $ \sum_{i = 1}^k M_i = \frac {k(n + 1)}{2} - \frac {k(k + 1)}{2} = \frac {k(n - k)}{2}$, and therefore $ (M_1,M_2, \cdots, M_n) \succ (m_1,m_2, \cdots, m_n)$. (Wlog $ \underline{m}$ is sorted decreasingly) Therefore by Karamata and the fact that $ x \to x^2$ is convex: $ f(\underline{m}) \le \sum_{i = 1}^n M_i^2 = \frac {n(n^2 - 1)}{12}$! Since $ \underline{m}$ is obtained by $ \Delta_i$ transformations from $ \underline{d}$, which is clearly valid, $ \underline{m}$ is valid. Hence there exists reals $ b_1,b_2,..,b_n$ such that $ b_i - a_i$ is an integer and: Let $ m_i = nb_i - \sum_{i = 1}^n b_i$. Then: $ n^2\sum_{1 \le i < j \le n} (b_i - b_j)^2 = \sum_{1 \le i < j \le n} (m_i - m_j)^2 =$ $ n\sum_{i = 1}^n m_i^2 - \left ( \sum_{i = 1}^n m_i^2 + 2\sum_{1 \le i < j \le n} m_im_j \right ) = n \sum_{i = 1}^n m_i^2 - \left ( \sum_{i = 1}^n m_i \right )^2 =$ $ n \sum_{i = 1}^n m_i^2 \le n\frac {n(n^2 - 1)}{12} = n^2 \frac {n^2 - 1}{12} \iff$. $ \sum_{1 \le i < j \le n} (b_i - b_j)^2 \le \frac {n^2 - 1}{12}$ So $ b_1,b_2,..,b_n$ fullfilles (b) and (c). Hence for some $ k \in \mathbb{Z}$ $ b_1 + k,b_2 + k, ..., b_n + k$ satisfies both (a) and (b), which is the conclusion

EthanWYX2009 showed me the following slick solution by using the probabilistic method (algebraic manipulations are omitted). Consider the elements $a_i \mod 1$ instead, and wlog assume $a_1 \le a_2 \le \cdots \le a_n$. Now, let $x$ be a variable uniformly distributed on the interval $[0, 1)$, and let $b_i(x) = \begin{cases} a_i & \text{ if } x \le a_i \\ 1 - a_i & \text{ if } x > a_i \end{cases}$ be a random variable, the expectation of the expression on the problem is \[\mathbb{E}[\sum_{1 \leq i < j \leq n} (b_i - b_j)^2] = \sum_{1 \leq i < j \leq n} \mathbb{E}[(b_i - b_j)^2] = \sum_{1 \leq i < j \leq n} (1 - a_i + a_j)(a_j - a_i) \]So now our task is to show that \[\sum_{1 \leq i < j \leq n} (1 - a_i + a_j)(a_j - a_i) \le \frac{n^2-1}{12}\]Notice how this is translation invariant, so we can also assume that $\sum a_i = 0$, so now \begin{align*} \sum_{1 \leq i < j \leq n} (1 - a_i + a_j)(a_j - a_i) &= \sum_{1 \leq i < j \leq n} (a_j - a_i) - \sum_{1 \leq i < j \leq n} (a_j - a_i)^2 \\ \sum_{i=1}^{n}(2i - 1 -n)a_i - n\sum_{i=1}^{n}a_i^2 &= n \sum_{i=1}^{n}\left(\frac{2i - 1 - n}{2n}\right)^2 - n\sum\text{something}^{2} \le \frac{n^2 - 1}{12} \end{align*}As desired