APMO

Observation from that $\mathrm{lcm}(2,3,4,5,6,7)=420$ is divisible by every integer less than or equal to $T=[\sqrt[3]{420}]$ and that $\mathrm{lcm}(2,3,4,5,6,7,8)=840$ is not divisible by $9=[\sqrt[3]{840}]$. One may guess $420$ is the required integer.

Let $N$ be the required integer and suppose $N>120$. Put $t=[\sqrt[3]{N}]$. Then

Since $t\ge T$, $\mathrm{lcm}(2,3,4,5,6,7)=420$ should divide $N$ and hence $N\ge 840$, which implies $t\ge 9$. But then $\mathrm{lcm}(2,3,4,5,6,7,8,9)=2520$ should divide $N$, which implies $t\ge 13=[\sqrt[3]{2520}]$.

Observe that any four consecutive integers are divisible by $8$ and that any two out of four consecutive integers have gcd either $1$, $2$ or $3$. So, we have $(t)(t-1)(t-2)(t-3)$ divides $6N$ and in particular,

$$t(t-1)(t-2)(t-3)\le 6N$$

From above follows

$$t(t-1)(t-2)(t-3)\le 6t(t^3+3t+3)$$

Since $t\ge 13$,

$$\frac{12}{t}+\frac{7}{t^2}+\frac{24}{t^3}<1$$

which is a contradiction.

Therefore, the largest such integer is $420$.