I recently solved this problem that was part of the excercises of a combinatorics book, it wasn't a hard problem, provided that I knew I had to use the probabilistic method, which the chapter was on. But I wanted to know if there is a non-probabilisic solution to this problem. Without using anything equivalent to the probabilistic method. The problem is as follows
Given $v_1,\cdots,v_n\in\mathbb R^n$ vectors such that $|v_i|=1$ for every $i$. Prove that there are numbers $a_i=\pm1$ such that
$$|a_1v_1+\cdots+a_nv_n|\leq \sqrt n$$
More over, show that we can also choose the $a_i$ such that
$$|a_1v_1+\cdots+a_nv_n|\geq \sqrt n$$
For any two vectors $u,v\in\mathbb R^n$, let $s=\operatorname{sign}\langle u,v\rangle$ (with the convention that $\operatorname{sign}(0)=1$). Then $$ \|u-sv\|^2 =\langle u-sv,u-sv\rangle =\|u\|^2-2s\langle u,v\rangle+\|v\|^2 \le\|u\|^2+\|v\|^2. $$ Similarly, $\|u+sv\|^2\ge\|u\|^2+\|v\|^2$.
The first part of your problem can therefore be solved by taking $a_1=1$ and $$ a_i=-\operatorname{sign}\langle a_1v_1+\cdots+a_{i-1}v_{i-1},\,v_i\rangle $$ for each $i\ge2$, while the second part is solved by taking $a_1=1$ and $$ a_i=+\operatorname{sign}\langle a_1v_1+\cdots+a_{i-1}v_{i-1},\,v_i\rangle $$ for each $i\ge2$.