10. Assume that $a_{i} \in\{1,-1\}$ for all $i=1,2, \cdots, 2013$. Find the least positive number of the following expression
$$
\sum_{1 \leq i<j \leq 2013} a_{i} a_{j} .
$$
10. Answer: 6
Solution. Note that
$$
\begin{aligned}
2 \sum_{1 \leq i<j \leq 2013} a_{i} a_{j} & =\left(a_{1}+a_{2}+\cdots+a_{2013}\right)^{2}-\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{2013}^{2}\right) \\
& =\left(a_{1}+a_{2}+\cdots+a_{2013}\right)^{2}-2013 .
\end{aligned}
$$

By the given condition, $a_{1}+a_{2}+\cdots+a_{2013}$ is an odd number between -2013 and 2013 inclusive.
Also note that the minimum positive integer of $x^{2}-2013$ for an integer $x$ is $45^{2}-2013=12$ when $x=45$ or -45 . As an illustration, $x=45$ can be achieved by taking $a_{1}=a_{2}=a_{3}=$ $\cdots=a_{45}=1$ and the others $a_{46}, a_{47}, \cdots, a_{2013}$ to consist of equal number of 1 's and -1 's. Thus the least value is $\frac{12}{2}=6$.