VMO

Firstly, we will prove that $\frac{f(x)}{x}$ is a constant. Assume that there exists $a,b\in (0,+\infty )$ such that $\frac{f(a)}{a}\ne \frac{f(b)}{b}$. Without loss of generality, we assume that $\frac{f(a)}{a}<\frac{f(b)}{b}$. By plugging $x=a,x=b$ into the relation, we get $$f\left(y+\frac{f(a)}{a}\right)=f\left(y+\frac{f(b)}{b}\right)=f(y)+1$$ for all positive real numbers $y$. Denote $K=\frac{f(b)}{b}-\frac{f(a)}{a}$, we obtain that $f(x)=f(x+K)$ for all $x>\frac{f(a)}{a}$. Furthermore, we have $$\begin{gathered} f\left(y+n\cdot \frac{f(a)}{a}\right)=f\left(y+(n-1)\cdot \frac{f(a)}{a}\right)+1 \\ =\cdots =f(y)+n>n. \end{gathered}$$ Thus $f(x)>n$ for all $x>n\cdot \frac{f(a)}{a}$. Now, for an arbitrary positive number $x$, choose $n=\lfloor f(x)\rfloor +2>f(x)$ and a positive integer $m$ such that $x+mK>n\cdot \frac{f(a)}{a}$, we have $f(x+mK)>n>f(x)=f(x+mK)$, which is a contradiction.

Hence, $\frac{f(x)}{x}$ is a constant. Denote $\frac{f(x)}{x}=c$ and replace $f(x)=cx$ in the original equation, one can find that $c=1$. Thus, $f(x)=x$ for all positive numbers $x$. $\square$