n_max=3. Rewrite as 3(x_i-x_j)^2>=(1+x_ix_j)^2. So if x_i>x_j, then sqrt(3)(x_i-x_j)>=1+x_ix_j. Since the system of inequalities are symmetric, then WLOG that x_1<x_2<...<x_n. Set x_i=tan(a_i), where 0<=a_i<90. Then a_1<a_2<...<a_n. Now we know that tan(a_i-a_j)=(x_i-x_j)/(1+x_ix_j)>=1/sqrt(3)=tan 30, so a_i-a_j>=30. So if n>=4, a_n>=a_(n-1)+30>=...>=a_(n-3)+90>=90, contradiction. So n<=3. Equality holds for n=3, when x_1=tan(a), x_2=tan(a+30), x_3=tan(a+60), for some 0<=a<30.

$(3x_i-x_j) (x_i-3x_j)\geq (1-x_ix_j)^2$ $\iff$ $\frac{x_j-x_i}{1+x_ix_j}\ge\frac{1}{\sqrt{3}}$ See also here Canadian Mathematical Olympiad - 1984