Question 3 Suppose the number $a$ has the property: for any four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, there always exist integers $k_{1}, k_{2}, k_{3},$ $k_{4}$, such that
$$
\sum_{1 \leqslant i<j \leqslant 4}\left(\left(x_{j}-k_{i}\right)-\left(x_{i}-k_{j}\right)\right)^{2} \leqslant a .
$$

Find the minimum value of such an $a$.
First, for $x_{i}(i=1,2,3,4)$, take $k_{i}=$ $\left[x_{i}\right]$, where $[x]$ represents the greatest integer not exceeding the real number $x$.
Let $\alpha_{i}=x_{i}-k_{i}$. Then $\alpha_{i} \in[0,1)$.
Let $\left\{\alpha_{1}^{\prime}, \alpha_{2}^{\prime}, \alpha_{3}^{\prime}, \alpha_{4}^{\prime}\right\}$ be a permutation of $\left\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right\}$ in ascending order, and denote
$$
\begin{array}{l}
\beta_{1}=\alpha_{2}^{\prime}-\alpha_{1}^{\prime}, \beta_{2}=\alpha_{3}^{\prime}-\alpha_{2}^{\prime}, \\
\beta_{3}=\alpha_{4}^{\prime}-\alpha_{3}^{\prime}, \beta_{4}=1-\alpha_{4}^{\prime}+\alpha_{1}^{\prime} .
\end{array}
$$

Then $S=\sum_{1 \leqslant i<j \leqslant 4}\left(\alpha_{i}-\alpha_{j}\right)^{2}$
$$
\begin{aligned}
= & \beta_{1}^{2}+\beta_{2}^{2}+\beta_{3}^{2}+\left(\beta_{1}+\beta_{2}\right)^{2}+ \\
& \left(\beta_{2}+\beta_{3}\right)^{2}+\left(\beta_{1}+\beta_{2}+\beta_{3}\right)^{2} .
\end{aligned}
$$

If the value of $k_{i}$ corresponding to $\alpha_{4}^{\prime}$ is increased by 1, while keeping the other $k_{i}$ values unchanged, then
$$
\begin{array}{l}
S=\beta_{4}^{2}+\beta_{1}^{2}+\beta_{2}^{2}+\left(\beta_{4}+\beta_{1}\right)^{2}+ \\
\left(\beta_{1}+\beta_{2}\right)^{2}+\left(\beta_{4}+\beta_{1}+\beta_{2}\right)^{2} . \\
\end{array}
$$

Similarly, $S$ can always be transformed into a sum of the form (2) using any three of $\left\{\beta_{1}, \beta_{2}, \beta_{3}, \beta_{4}\right\}$, and appropriate values of $k_{1}, k_{2}, k_{3}, k_{4}$ can be chosen so that the expression in (2) is missing the largest $\beta_{i}$. Without loss of generality, assume $\beta_{4}$ is the largest, and $S$ is expressed as in (2).
In (2), $\beta_{1}+\beta_{2}+\beta_{3}=1-\beta_{4}$, and clearly $\beta_{4} \geqslant \frac{1}{4}$.
When estimating the value of $S$, assume
$$
\beta_{2}=\max \left\{\beta_{1}, \beta_{2}, \beta_{3}\right\} .
$$

From (2), it is known that $S$ decreases in other cases.
If $\beta_{4} \geqslant \frac{1}{2}$, then $S \leqslant 4\left(1-\beta_{4}\right)^{2} \leqslant 1$.
Now consider $\frac{1}{4} \leqslant \beta_{4} \leqslant \frac{1}{2}$. In this case,
$$
\begin{array}{l} 
S= 3 \beta_{2}^{2}+2\left(\beta_{1}^{2}+\beta_{3}^{2}\right)+2 \beta_{2}\left(\beta_{1}+\beta_{3}\right)+\left(1-\beta_{4}\right)^{2} \\
= 3 \beta_{2}^{2}+\left(\beta_{1}-\beta_{2}\right)^{2}+\left(\beta_{1}+\beta_{3}\right)^{2}+ \\
2 \beta_{2}\left(\beta_{1}+\beta_{3}\right)+\left(1-\beta_{4}\right)^{2} \\
= 2 \beta_{2}^{2}+\left(\beta_{1}-\beta_{3}\right)^{2}+2\left(1-\beta_{4}\right)^{2} \\
\leqslant 2\left(\beta_{4}^{2}+\left(1-\beta_{4}\right)^{2}\right)+\left(\beta_{1}-\beta_{3}\right)^{2} \\
= 2\left(\beta_{4}^{2}+\left(1-\beta_{4}\right)^{2}\right)+\left(1-2 \beta_{4}\right)^{2} . \\
\text { When } \frac{1}{3} \leqslant \beta_{4} \leqslant \frac{1}{2} \text {, } \\
S \leqslant \max \left\{\frac{11}{9}, 1\right\}=\frac{11}{9} ; \\
\text { When } \frac{1}{4} \leqslant \beta_{4} \leqslant \frac{1}{3} \text {, } \\
S \leqslant \max \left\{\frac{11}{9}, \frac{10}{8}\right\}=\frac{10}{8}=\frac{5}{4} .
\end{array}
$$

When $\frac{1}{3} \leqslant \beta_{4} \leqslant \frac{1}{2}$,
$S \leqslant \max \left\{\frac{11}{9}, 1\right\}=\frac{11}{9}$;
When $\frac{1}{4} \leqslant \beta_{4} \leqslant \frac{1}{3}$,
$S \leqslant \max \left\{\frac{11}{9}, \frac{10}{8}\right\}=\frac{10}{8}=\frac{5}{4}$.
In summary, $S \leqslant \frac{5}{4}$.
It is easy to verify that when $x_{1}=0, x_{2}=\frac{1}{4}, x_{3}=\frac{1}{2}, x_{4}=\frac{3}{4}$, $S=\frac{5}{4}$.
Therefore, the minimum value sought is $\frac{5}{4}$.