We claim that the answer is $\frac{n}{\sqrt3}$ if $n \le 2$ and $n - 1$ for $n \ge 3.$ For $n = 1$ this is obvious. For $n = 2$, it suffices to show that if $\alpha, \beta \in (0, \frac{\pi}{2})$ so that $\cos \alpha + \cos \beta > \frac{2}{\sqrt3}$, we have $\tan \alpha \tan \beta < 2.$ Equivalently, if $a, b \in (0, 1)$ so that $a+b > \frac{2}{\sqrt3}$, we want to show that $(\frac{1}{a^2} - 1)(\frac{1}{b^2} - 1)<2.$ To show this, it suffices to show that for $0 < a < 1$, we have: $$(\frac{1}{a^2} - 1)(\frac{1}{(2 - \sqrt3 - a)^2} - 1) \le 2.$$ Checking this is simple by taking a derivative, and the details are left to the reader as an exercise. Let's show that for $n = 3$, if $\alpha, \beta, \gamma \in (0, \frac{\pi}{2})$ so that $\cos \alpha + \cos \beta + \cos \gamma >2$, then $\tan \alpha \tan \beta \tan \gamma < 2 \sqrt 2.$ This would suffice. Equivalently, if $a, b, c$ are real numbers with $a+b+c=2$, we wish to show that $$(\frac{1}{a^2}-1)(\frac{1}{b^2}-1)(\frac{1}{c^2}-1) \le 2.$$ Observe that $\ln (\frac{1}{x^2}-1)$ has only one inflection point on $(0, 1)$, and therefore by $n-1$ EV we can assume that $a = b.$ We now have to show that: $$(\frac{1}{a^2}-1)^2(\frac{1}{(2-2a)^2}-1) \le 2,$$ for all $a \in (\frac12, 1).$ This is again a straightforward bash, and left as an exercise to the reader. Now, we will solve it for $n > 3$ by induction. Our base case will be $n =3.$ Observe that there must be an $x_i$ with $\tan x_i \le \sqrt 2$ (e.g. just look at the smallest one). Remove this $x_i$, and notice that the product of the remaining $x_j$'s is at least $2^{\frac{n-1}{2}}$. Hence, the inductive hypothesis implies that the sum of the $\cos$'s of the remaining $x_j$'s is at most $n-2.$ Now, simply noting that $\cos x_i \le 1$ completes the induction. The induction is complete, and so we've shown that $\lambda \le n-1$ when $n \ge 3$ and $\lambda \le \frac{n}{\sqrt3}$ for $n = 1, 2.$ We'll now show that these are optimal. For $n \ge 3$, simply let $x_1, x_2, \cdots, x_{n-1}$ be $\epsilon$ for small $\epsilon$ and then define $x_n$ accordingly. For $n = 1, 2$, simply let all the $x_i$'s be equal. These constructions show that our obtained upper bounds for $\lambda$ are actually optimal, and so in conclusion the answer is $\lambda = \frac{n}{\sqrt3}$ for $n = 1, 2$ and $\lambda = n-1$ otherwise. $\square$

Another solution for $n=2$. If $n=2$, then $\frac 12\left(\cos (x_1-x_2)-\cos (x_1+x_2)\right)=\sin x_1\sin x_2=2\cos x_1\cos x_2 =\cos (x_1+x_2)+\cos (x_1-x_2)$. So we can get $\cos (x_1-x_2)+3\cos (x_1+x_2)=0$. Let $x=\cos\frac{x_1+x_2}{2}$, $y=\cos \frac{x_1-x_2}{2}$. Then $\cos (x_1-x_2)+3\cos (x_1+x_2)=2y^2-1+3(2x^2-1)=2y^2+6x^2-4=0$, $y^2+3x^2=2$. Therefore $\cos x_1+\cos x_2=2\cos\frac{x_1+x_2}{2}\cos\frac{x_1-x_2}{2}=2xy\leqslant\frac{2\sqrt 3}{3}$.