Belarusian Mathematical Olympiad

Answer: $n=2$.

Note that if $n=2$, the sequence has the form $a_k=k(k+1)$ since consecutive numbers are always coprime. It is clear that $k(k+1)<(k+1)(k+2)$, so the sequence is increasing.

First solution. Let us show that if $n\ge 3$ the sequence is not increasing from any number. Choose $k=np$ where $p$ is an arbitrary prime number greater than $n$. All numbers $np+1,np+2,\dots ,np+n-1$ are not divisible by $p$ and there is at least one even among them. The number $n(p+1)$ is also not divisible by $p$ and $p+1$ is even. Hence $$\begin{gathered} a_k=p\cdot \text{lcm}(n,np+1,np+2,\dots ,np+n-1), \\ {a}_{k+1}\le \frac{p+1}{2}\cdot \text{lcm}(n,np+1,np+2,\dots ,np+n-1). \end{gathered}$$ Therefore $a_k>a_{k+1}$ for $k$ large enough and the sequence $a_k$ is not increasing from any number.

Second solution. We will show how else one can prove that for $n\ge 3$ the sequence is not increasing from any moment. Suppose that $a_k$ is increasing from number $k_0$ then for all $k\ge k_0$ holds $a_{k+1}\ge a_k$. Consider $k=m!-n$ where $m>\max (n!+n,k_0)$.

Denote $\text{lcm}(k+1,\dots ,k+n-1)$ by $N$. Then the inequality $a_{k+1}\ge a_k$ is equivalent to $\text{lcm}(N,m!)>\text{lcm}(m!-n,N)$. Using the well known equality $\text{lcm}(a,b)\cdot \text{gcd}(a,b)=a\cdot b$ we obtain the equivalence: $$\text{lcm}(N,m!)>\text{lcm}(m!-n,N)⟺\frac{N\cdot m!}{\text{gcd}(N,m!)}>\frac{(m!-n)\cdot N}{\text{gcd}(m!-n,N)},$$ or, after equivalent transformations, $$n\cdot \text{gcd}(N,m!)>m!\cdot (\text{gcd}(N,m!)-\text{gcd}(m!-n,N)).(1)$$ It is clear that $\text{gcd}(m!,m!-l)=\text{gcd}(m!,l)=l$ for all $l$ from $1$ to $n-1$, hence $(n-1)!\ge \text{gcd}(N,m!)\ge \text{lcm}(1,2,\dots ,n-1)$. Moreover, since $m!$ is divisible by $l$, the number $\text{gcd}(m!-n,m!-n+l)=\text{gcd}(n,n-l)$ is a divisor of $n$. So the number $\text{gcd}(m!-n,N)$ is a divisor of $n$ too. Indeed ${\nu }_p(N)=\max ({\nu }_p(m!-n+1),\dots ,{\nu }_p(m!-n+(n-1)))$ for any prime divisor $p$ of $N$. From obtained inequalities and (1) it follows that $$n\cdot (n-1)!>m!\cdot (\text{lcm}(1,2,\dots ,n-1)-n).(2)$$ Since $\text{lcm}(1,2,\dots ,n-1)\ge (n-2)(n-1)$, for $n\ge 4$ the right hand side of (2) is not less than $m!$ whence $n!>m!$, a contradiction. If $n=3$ then $\text{gcd}(N,m!)=\text{gcd}(m!-2,m!)=2$. But $$\text{gcd}(m!-n,N)=\text{gcd}(m!-3,(m!-2)(m!-1))=1,$$ so the inequality (1) will become $6>m!$ which is wrong.