I was trying to solve Putnam's B6 from 1995:
"For any $a>0$,set $S(a)=\{\lfloor{na}\rfloor|n\in \mathbb{N}\}$. Show that there are no three positive reals $a,b,c$ such that $$S(a)\cap S(b)=S(b)\cap S(c)=S(c)\cap S(a)=\emptyset$$ $$S(a)\cup S(b)\cup S(c)=\mathbb{N}."$$
I looked for the natural density of each set, from where I deduced that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ (one can see that by using the fact that the natural density of $S(k)$ is $\frac{1}{k}$ and that $S(a),S(b)$ and $S(c)$ are disjoint), and without much manipulation one can reduce the problem to proving the following statement:
"If $x,y,z>0$ are distinct irrational numbers such that $x+y+z=1$, then there exists a positive integer $n$ such that $$\{nx\}+\{ny\}+\{nz\}\ne1."$$
I attempted proving this using Kronecker's Theorem, taking a sequence of $n_k$ such that $\{n_kx\}>1-\epsilon$, for some fixed small $\epsilon>0$, and trying to replicate the same Pigeonhole idea that proves the said theorem to prove that $\{n_ky\}+\{n_kz\}>\epsilon$ for some $k$. Unfortunately, this idea seems to not even get close to solve the said problem.
Furthermore, trying to do same and make the sum smaller than one seems to end up in the same problem. Also, this idea doesn't seem to be the one that proves the statement since it's false for two irrationals: For example, take $x=\sqrt{2}$ and $y=2-\sqrt{2}$ and the sum $\{nx\}+\{ny\}=1$, for all positive integer $n$, and therefore we should use the fact that we have $3$ irrationals.
I haven't yet had a promissing idea nor any memorable advance into proving this statement. I've tried to imagine why some more obvious counter-examples such as $x=1-\frac{\sqrt{2}}{2}$ and $y=z=\frac{\sqrt{2}}{4}$ don't work, but when two numbers have the same fractional part, one can just use Kronecker's Theorem.
I've read that for an irrational $\alpha>0$, $(\{n\alpha\})_{n\geq1}$ is uniformely distributed in $(0,1)$ (https://mathoverflow.net/questions/131018/sequences-equidistributed-modulo-1), which would imply that
$$\lim_{n\to\infty}\frac{1}{n}\sum_{1\leq k\leq n}\{k\alpha\}=\frac{1}{2},$$
which seems to prove the following statement, but I couldn't find a prove of this without complex analysis (https://www-2.dc.uba.ar/staff/becher/papers/koksma.pdf), which doesn't seem suitable (even though one could argument that it is) for an undergrad olympiad and even more to an olympiad such as Putnam, which has, in general, completely elementary solutions. If there is a proof for this statement without complex analysis, I'd love to see it.