Oh,can somebody give a solution?I am curious about this problem. I think this one is not elementary at all.Maybe we can use Weil Theorem.

I like this problem a lot... Some trivial progress: We have - $ na + nb - \{na\} - \{nb\} = nc + nd - \{nc\} - \{nd\} \implies$ $ n(a + b - c - d) = \{na\} + \{nb\} - \{nc\} - \{nd\}$ If $ a + b \neq c + d$, then the LHS is unbounded while the RHS is bounded. Thus, $ a + b = c + d$. 2 possible ways to continue: I read about the fractional part function and the following identity holds: $ \{x\} = \frac {1}{2} - \frac {1}{\pi} \sum_{k = 1}^\infty \frac {\sin(2 \pi k x)} {k}$ (from Wikipedia), this is the Fourier series of $ \{x\}$. Thus, the relation $ \{na\} + \{nb\} = \{nc\} + \{nd\}$ is equivalent to: $ \sum_{k = 1}^\infty \frac {\sin(2 \pi k na) + \sin(2 \pi k nb)} {k} = \sum_{k = 1}^\infty \frac {\sin(2 \pi k nc) + \sin(2 \pi k nd)} {k} \forall n \in \mathbb{N}$ Or: $ \sum_{k = 1}^\infty \frac {\sin(2 \pi k na) + \sin(2 \pi k nb) - (\sin(2 \pi k nc) + \sin(2 \pi k nd))} {k} = 0$ Notice that when $ a + b = c + d$, we have: $ \sin(a) + \sin(b) - \sin(c) - \sin(d) = - 4\sin(\frac {a + b}{2})\sin(\frac {a + c - b - d}{2})\sin(\frac {a + d - b - c}{2})$ So we have: $ \sum_{k = 1}^\infty \frac {\sin(\pi nk(a + b))\sin(2\pi nk(a - d))\sin(2\pi nk(a - c))} {k} = 0$ But $ a + b, a - c$ or $ a - d$ are integers iff the numerator of the last fraction is 0 for all $ n$ and $ k$! I think it may lead to a solution... 2nd possible way - the last nice identity here: http://mathworld.wolfram.com/FloorFunction.html

Very late, but I wonder what Weil thm is. Could you give a link about it?