1 Autumn tournament

For $k=1$ and $k=2$, the required polynomials are $f(x)=x+1$ and $f(x)=(x+1)(x+2)$, respectively. Let $k\ge 3$ and assume that such a polynomial $f(x)$ exists. For any prime number $p$, its degree in $\mathrm{lcm}(n+1,n+2,\dots ,n+k)$ is $\max \{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_{k-1}\}$, where ${\alpha }_i$ is the power of $p$ in the canonical representation of $n+i$, $i=1,\dots ,k$. If this is, for example, ${\alpha }_s$, then it would be obtained if we take $$\frac{(n+1)(n+2)\cdots (n+k)}{p^{{\alpha }_1}{p}^{{\alpha }_2}\cdots p^{{\alpha }_{s-1}}{p}^{{\alpha }_{s+1}}\cdots p^{{\alpha }_k}},$$ it being clear that the powers of $p$ in the denominator are divisors of ${\prod }_{1\le i\ne s\le k}(s-i)$. Therefore $$\mathrm{lcm}(n+1,n+2,\dots ,n+k)=\frac{(n+1)(n+2)\dots (n+k)}{C_n},(1)$$ where $C_n$ is a divisor of ${\prod }_{1\le i<j\le k}(j-i)$. Since $C_n$ can take a finite number of possible values, there will be a natural number $C$ such that for infinitely many $n$, $f(n)=\frac{(n+1)(n+2)\dots (n+k)}{C}$. So for infinitely many $x$, $f(x)=\frac{(x+1)(x+2)\dots (x+k)}{C}$, whence $$f(x)=\frac{(x+1)(x+2)\dots (x+k)}{C},\forall x\in \mathbb{R}.$$ Therefore $$\mathrm{lcm}(n+1,n+2,\dots ,n+k)=\frac{(n+1)(n+2)\dots (n+k)}{C},\text{for all }n\in \mathbb{N}.$$ Let's assume this is possible. Let's choose a prime $p<k$ such that $p$ does not divide $k$. Let $n+k+1=p^m$ for sufficiently large $m$. From the above formula we have $$\frac{\mathrm{lcm}(n+2,n+3,\dots ,n+k+1)}{\mathrm{lcm}(n+1,n+2,\dots ,n+k)}=\frac{n+k+1}{n+1}.(2)$$ The degree of $p$ in the numerator of the left side is $m$ and in the denominator – at least 1, while the degree of $p$ in the numerator of the right side is $m$ and in the denominator – 0. We derive a contradiction! Therefore, the assumption is wrong and for $k\ge 3$ there does not exist a polynomial with the desired property. $\square$