Let $s_k$ be the $lcm$ of the numbers $1$ to $k$. Is $s_{28}$ the largest highly composite such number? A highly composite number is a natural number $n\ge1$, which has more divisors than any smaller number $m\ge 1$.
Condider the numbers $s_k:=lcm(1,\cdots, k)$
Checking the $1000$ entries in OEIS, I found the following highly composite numbers 
$[1, 2, 6, 12, 60, 840, 2520, 27720, 720720, 80313433200]$
So, $s_{28}$ seems to be the largest such number.
This question is closely related to the question about the largest highly composite factorial, which has apparantly been solved :
Is $\ 7!=5040\ $ the largest highly composite factorial?
I think, similar arguments will show that $s_{28}$ is the largest highly composte number, being the $lcm$ of the numbers from $1$ to $k$. Can anyone prove it ?
 A: The idea is that the numbers $s_k$ don't have enough small factors to make the highly composite list.  $s_{28}=80313433200=2^4\times3^3\times 5^2\times 7\times 11\times 13\times 17\times 19\times 23$  When we multiply by $29$ to get $s_{29}$ we double the number of factors, but we can multiply by $28=2^2\times 7$ to increase the number of factors by $\frac 75 \cdot \frac 32=2.1$ times, so $s_{29}, s_{30},$ and $s_{31}$ cannot be highly composite as $28s_{28} \lt s_{29}$ and has more factors. $s_{32}$ cannot be highly composite because $s_{32}=62s_{29}$ and has a factor $2 \cdot \frac 65=2.4$ more factors, but $56s_{29}$ has a factor $\frac 85 \cdot \frac 32 =2.4$.  After $s_{32}$ we have another $2$ in the factorization, but we can increase the number of factors by a factor $\frac 76 \cdot \frac 54 \cdot \frac 43=2$ by multiplying by $30$, which is better than the doubling we get by multiplying by a new prime like $37$.  By now the $s$ numbers are huge compared to highly composites with the same number of factors.  I haven't found a nice rule to find the multipliers to be used as te numbers get higher.
