Let $m^3 \leq n < (m+1)^3$; then we need $\dfrac {m(m-1)(m-2)(m-3)} {12} \leq \textrm{lcm}[1,2,\ldots,m-1,m] \leq n \leq m^3 + 3m^2 + 3m$. This implies $m\leq 19$, and now it's a simple matter of manually checking this small finite number of cases. But $\textrm{lcm}[1,2,\ldots,m] \geq \textrm{lcm}[1,2,\ldots,11] = 27720 > 20^3$ for $m\geq 11$; $\textrm{lcm}[1,2,\ldots,10] = 2520 > 11^3$; $\textrm{lcm}[1,2,\ldots,8] = 840 > 9^3$; $\textrm{lcm}[1,2,\ldots,7] = 420 < 8^3$. Therefore the largest such is $n=420$, with $7^3 < 420 < 8^3$.

why $ \frac{m(m-1)(m-2)(m-3)}{12}\leq\textrm{lcm}[1,2,\ldots,m-1,m] $ ?

Because $\gcd(m,m-1) = 1$, $\gcd(m,m-2) \leq 2$, $\gcd(m,m-3) \leq 3$; $\gcd(m-1,m-2) = 1$, $\gcd(m-1,m-3) \leq 2$; $\gcd(m-2,m-3) = 1$.

ok thanks

Obviously there exists an positive integer a such that $a^3<x<(a+1)^3$. $x=a^3+k$ where $0<k<3a^2+3a+1$. If $a>8, a(a-1)|k\implies k\in{a^2-2a,2a^2-3a,3a^2-4a}$. $a-2|k$ gives $a\in$ $\{10$,$12$,$14 \} $ which gives no solution. If $k<9$, $k=8$ gives $x\ge 840$ which is not a solution. For $k=7$ we have $x=420$

I got the answer

Notice the functions $f(n)=lcm(1,2…n)$ and $g(n)=n^3$ checking for $f(1),f(2)…f(7)$ we see that there exists $k\in Z :k^3<n:k\not | n$ so largest number is $f(7)=420=1.420=2.210=3.140=4.105=5.84=6.70=7.60$