As far as I know, this is not an IMO question... certainly not from 1987. But a proof of Beatty's theorem can be found here on page 6-7.

Peter wrote: As far as I know, this is not an IMO question... certainly not from 1987. But a proof of Beatty's theorem can be found here on page 6-7. Yeah, you're right. It is not from IMO problems.

Proof of Beaty's theorem from Cours d'Arithmetique I Part,par P.Bornsztein, X. Caruso, P. Nolin, M. Tibouchi, 2004. My translation. Suppose $ \alpha$ and $ \beta$ be two irrational numbers such that $ \frac{1}{ \alpha} + \frac {1}{\beta} = 1$. Let $ k$ be a strictly positive integer . $ k$ will be an element of $ S_\alpha$ if and only if there exists an integer $ n$ such that $ n \alpha - 1 < k < n \alpha$. The RHS of inequality is strict one, since $ \alpha$ is supposed to be irrational. The equation can be transformed into the following one: $ \frac{k}{\alpha} < n < \frac{k}{\alpha} + \frac{1}{\alpha}$. That means , the following. Namely $ k$ would be element of $ S_ {\alpha}$ if and only if the interval $ ] \frac{k}{\alpha}, \frac{k}{\alpha} + \frac{1}{\alpha} [$ contains an integer. The same: $ k$ would be element of $ S_ {\beta }$ if and only if the interval $ ]\frac{k}{\beta}, \frac{k}{\beta} + \frac{1}{\beta}[$ contains an integer . The lenght of the interval $ ]\frac {k}{\alpha}, \frac{k}{\alpha} + 1[$ is equal to $ 1$ and both of its end-points are irrational , so that is why it consists one and only one integer $ n$. If $ n < \frac{k}{\alpha} + \frac{1}{\alpha}$ then $ k \in S_{\alpha}$. But, if not we will have another inequality, namely: $ \frac{k}{\alpha}+\frac{1}{\alpha} < n < \frac{k}{\alpha} +1$. The LHS of this inequality is strict one since $ \frac {k+1}{\alpha}$ is irrational. Since $ \frac{k}{\alpha} = k - \frac{k}{\beta}$ we have $ \frac{k}{\beta} < k + 1 - n < \frac {k}{\beta} + \frac {1}{\beta}$ so $ k \in S_{\beta}$. If $ k$ would be simultaneously element, both of $ S_{\alpha}$ and $ S_{\beta}$, then it would be an integer in the interval $ ]\frac{k}{\alpha},\frac{k}{\alpha} + \frac{1}{\alpha}[$ and in the interval $ ]\frac{k}{\beta},\frac{k}{\beta} + \frac{1}{\beta}[$. So, in the same way as before it would be two elements in the interval $ ]\frac{k}{\alpha}, \frac {k}{\alpha} +1[$, which is impossible. Conversely, suppose that $ S_{\alpha}$, $ S_{\beta}$ partition the set $ N$. Let us consider a strictly positive integer $ k$. So, there exist $ [\frac {k}{\alpha}]$ integers in the set $ {1,...,k}$ being elements of $ S_{\alpha}$. Similarly , there exist $ [\frac {k}{\beta}]$ integers in $ {1,...,k}$ being elements of $ S_{\beta}$. Because of the partition it appears that, $ [\frac{k}{\alpha}] + [\frac{k}{\beta}] = k$ for all $ k$. Let $ k \rightarrow \infty$, so we have $ \frac{1}{\alpha} + \frac{1}{\beta}= 1$ and that proves the second condition. Suppose now, that $ \alpha$ be rational. So, the same for $ \beta$ after the previous relations. Let us write $ \alpha = \frac{a}{b}$ and $ \beta = \frac{c}{d}$. The integer $ a\cdot c$ is element of $ S_{\alpha}$ (taking $ n = b\cdot c$) and equally element of $ S_{\beta}$ (taking $ n = a \cdot d$ ) which is the contradiction.

let set $A= \lfloor\alpha\rfloor,\lfloor 2\alpha\rfloor,\lfloor 3\alpha\rfloor,\cdots$ and set $B= \lfloor\beta\rfloor,\lfloor 2\beta\rfloor,\lfloor 3\beta\rfloor,\cdots$ note that no two member are equal in $A$ and no two pair of members are equal in $B$ because $\alpha,\beta>1$ then we prove that there are exactly $n-1$ members of $A$ and $B$ that are less that $n$. let $x$ be the greatest natural number for which $\lfloor x\alpha\rfloor <= n-1 \implies x=\lfloor n\alpha^{-1}\rfloor $ and we define $y$ the same way: $\lfloor y\beta\rfloor <= n-1 \implies y=\lfloor n\beta^{-1}\rfloor $ thereupon easily we get that $x+y=\lfloor n\alpha^{-1}\rfloor+\lfloor n\beta^{-1}\rfloor=n-1$ owing to the fact that $\beta^{-1}+\alpha^{-1}=1$ . so we know there are exactly $n-1$ members of $A$ and $B$ that less that $n$ and there are exactly $n$ members of $A$ and $B$ that less that $n+1$ for any $n>1$ which mean there's exactly one member in $A$ and $B$ equal to $n$ we any natural number greater than $2$ and it's totally obvious that $1$ is in $A$ or $B$ and proving that happens only for one member of $A$ and $B$ is straightforward