Romanian Mathematical Olympiad

Let $N$ be an arbitrary positive integer. There exists a least $k\in \mathbb{N}$ such that $x_N<\frac{1}{N}+\frac{1}{N+1}+\cdots +\frac{1}{N+k}$, since the harmonic series is divergent. Then $x_{N+k}=x_N-\left(\frac{1}{N}+\cdots +\frac{1}{N+k-1}\right)<\frac{1}{N+k}\le \frac{1}{N}$. Also $x_n<\frac{1}{N}$ for $n\ge N+k$. The proof is by simple induction, since $x_{n+1}=\left|x_n-\frac{1}{n}\right|$, and if $x_n>\frac{1}{n}$, then $x_{n+1}=x_n-\frac{1}{n}<x_n<\frac{1}{N}$, while if $x_n\le \frac{1}{n}$, then $x_{n+1}=\frac{1}{n}-x_n\le \frac{1}{n}<\frac{1}{N}$.

Therefore for any $N$ there exists $N_1$ such that $0\le x_n<\frac{1}{N}$ for $n\ge N_1$, which means the sequence converges to $0$.