We can prove by induction that $\text{lcm}(n,n+k)=\frac{n(n+k)}{k}$ for $k=1,...,35$. Indeed, we have $\text{lcm}(n,n+k)=\frac{n(n+k)}{\text{gcd}(n,n+k)}<\frac{n(n+k-1)}{k-1}$, which is only possible if $\text{gcd}(n,n+k)=k$. Thus we have $\text{gcd}(n,n+4)=4\Rightarrow 4|n$ and $\text{gcd}(n,n+9)=9 \Rightarrow 9|n$ and it follows that $\text{gcd}(n,n+36)=36$. Thus $\text{lcm}(n,n+35)=\frac{n(n+35)}{35} >\frac{n(n+36)}{36}=\text{lcm}(n,n+36)$.

Cute one. Here $\gcd(a,b)=(a,b).$ Note that $[a,b]=\frac{a\cdot b}{(a,b)}.$ Also note that $(n,n+k)\le k.$ So, $$\frac{n\cdot (n+1)}{(n,n+1)}= ~~\frac{n\cdot (n+1)}{1}=~~ n\cdot (n+1).$$Now if $2\nmid n,$ then $$(n,n+2)=1 \implies [n, n + 2] = n\cdot (n+2) > [n,n+1]. $$Not possible. Hence $2|n.$ And $[n, n + 2]= \frac{n\cdot (n+2)}{2}.$ Similarly, if $$3\nmid n,$$then $$(n,n+3)=1\implies [n, n + 3] = n\cdot (n+3)> [n,n+1]. $$Not possible. Hence $3|n.$ And $[n, n + 3]= \frac{n\cdot (n+3)}{3}.$ Lemma :- So here, we can say that for any prime $p<35, ~~[n,n+p]=\frac{n\cdot (n+p)}{p}.$ If not then $(n,n+p)=1.$ But then $$[n,n+p]=n\cdot (n+p)> n\cdot (n+1)=[n,n+1].$$ Also we must have $(n,n+4)=4.$ If not then $(n,n+4)=2. $ (Since $2|n.$) But then $$\frac{n\cdot (n+2)}{2} <\frac{n\cdot (n+4)}{2}.$$Not possible. Hence $4|n.$ Similarly, we must have $(n,n+9)=9.$ If not then $(n,n+9)=3. $ (Since $3|n.$) But then $$\frac{n\cdot (n+3)}{3} >\frac{n\cdot (n+9)}{3}.$$Not possible. Hence $9|n.$ Hence $36|n$ and $[n,n+36]=\frac{n\cdot (n+36)}{36}.$ Consider $[n, n + 35] ,~~ [n,n + 36].$ By lemma we get $$5|n, ~~7|n \implies 35|n \implies [n, n + 35]=\frac{n\cdot (n+35)}{35}. $$So $$[n, n + 35]=\frac{n\cdot (n+35)}{35} , ~~ [n,n+36]=\frac{n\cdot (n+36)}{36}.$$ But clearly as $\frac{n}{35}>\frac{n}{36},$ $$\frac{n\cdot (n+35)}{35} >\frac{n\cdot (n+36)}{36}$$. And we are done!