It's strange because I got the result: $\left\lceil \frac{s^{2}(1-\lambda)^{2}}{n+n\alpha^{2}-2s\alpha}\right\rceil$ where $\alpha=\frac{s\lambda}{n}$

This has been sitting without a solution for more than a decade. Here is a probabilistic proof. I'll use Cantelli's inequality: Thm (Cantelli): Let $X$ be a random variable, with standard deviation $\sigma$. Then, for any $t<0$: $$ \mathbb{P}(X-\mathbb{E}[X]>t)\geqslant 1-\frac{\sigma^2}{\sigma^2+t^2}. $$Now, equipped with this, let $X$ be a random variable uniformly distributed on $S=\{x_1,x_2,\dots,x_n\}$, that is, $\mathbb{P}(X=x_i)=1/n$ for each $i $. Note that, $\mathbb{E}[X^2]=1$, and $\mathbb{E}[X]=s/n$; and therefore, $\sigma^2 = 1-\frac{s^2}{n^2}$, where $\sigma$ is the standard deviation of $X$. With this, observe the following chain: $$ \mathbb{P}(X\geqslant \frac{\lambda s}{n}) = \mathbb{P}(X\geqslant \lambda \mathbb{E}[X]) = \mathbb{P}(X-\mathbb{E}[X]\geqslant (\lambda-1)\mathbb{E}[X]) \geqslant \frac{(1-\lambda)^2\mathbb{E}[X]^2}{(1-\lambda)^2\mathbb{E}[X]^2+\sigma^2}. $$Now, inserting $\mathbb{E}[X]=s/n$, and $\sigma^2 = 1-s^2/n^2$, we have: $$ \mathbb{P}(X\geqslant \frac{\lambda s}{n}) \geqslant \frac{(1-\lambda)^2 s^2}{(1-\lambda)^2s^2 + n^2-s^2}>\frac{(1-\lambda)^2s^2}{n^2}. $$Thus, the 'number' of $x_i$'s above this level is nothing but $n$ times the lower bound (due to the fact that the probability mass is $1/n$), that is, there is at least $>\frac{(1-\lambda)^2s^2}{n}$ values of $x_i$ with the desired condition, from which the ceil is obtained.