For $n=2$ we have $a_k=[k,k+1]=k(k+1)$ which increases. If $n\ge 3$ we let $p$ be a prime factor of $n-1$ and we choose $k>n$ with $k\equiv -n(p)$ and $k\equiv 1(q)$ when $q<n,q\not=p$. Then $(k,(n-1)!)=1$ and we can send $k\rightarrow \infty$. We have $$a_{k+1}=[k+1,\cdots,k+n]\le [k+1,\cdots,k+n-1] {{k+n}\over p}\le [k+1,\cdots,k+n-1]{{k+n}\over 2}<k[k+1,\cdots,k+n-1]=[k,\cdots,k+n-1]=a_k$$Hence $a_k$ is not ultimately increasing.

Let's select any prime $p>n$. Let $k=pn$, and $s=\text{lcm}(pn+1,\cdots , (p+1)n-1)$. Then we have $a_k=\text{lcm}(pn, s)$ and $a_{k+1}=\text{lcm}((p+1)n, s)$. Now, if $n\notin \mathbb{P}$, then $n\mid s$, therefore $a_k=\text{lcm}(p,s), a_{k+1}=\text{lcm}(p+1,s)$. However, since $\text{gcd}(p, np+i)=1$ for $i=1,2\cdots, n-1$, we have $\text{lcm}(p,s)=ps$. And $2\mid p+1$, therefore $\text{lcm}(p+1,s)\leq \frac{(p+1)(s)}{2}<ps$. So we get that $a_k<a_k+1$. If $n \in \mathbb{P}$ we get that $\text{gcd}(s,n)=1,$ therefore $\text{gcd}(pn,s)=\text{gcd}(p,s)n$ and $\text{gcd}((p+1)n,s)=\text{gcd}((p+1),s)n$. Further solution is analogical. Finally, we get the only possible $n=2$ which obviously works.