What is the sum of divisors of binomial&factorial $$\sum_{m|\frac{n!}{i!(n-i)!}} m$$ 
Perhaps a good start is $$\sum_{m|n!} m$$ When seeing this last sum or also this one $$\sum_{m|lcm(1,2,...,n)} m$$ I sort of want to use $(n+1)n\over2$ somehow. 
 A: Since there has not be an answer I post an ugly looking formula and hopefully edited the question to something a bit more interesting. This is not the answer I desire. Maybe you have suggestions.
I use Euler's short notation: $\int a=\sum_{m|a}m$ and the well known properties: $\int p^n=1+p+\dots+p^n=\frac{p^{n+1}-1}{p-1}$ when p is prime, $\int ab=\int a \int b$ when $gcd(a,b)=1$. When $p_i$ is the i'th prime one can write $$n!=\prod_{i=1}^{\pi(n)}p_i^{\sum_{k=1}^{\lfloor {ln(n)/ln(p_i)}\rfloor} \lfloor {n\over p_i^k}\rfloor }=\prod_{i\geq1} p_i^{\sum_{k\geq 1} \lfloor {n\over p_i^k}\rfloor }$$ The first equality uses the least number of terms but I use the second since it looks nicer. Now according to this $$\int n! = \int \prod_{i\geq1} p_i^{\sum_{k\geq 1} \lfloor {n\over p_i^k}\rfloor }=\prod_{i\geq1} \int p_i^{\sum_{k\geq 1} \lfloor {n\over p_i^k}\rfloor }=\prod_{i\geq1} \frac{p_i^{\sum_{k\geq 1} \lfloor {n\over p_i^k}\rfloor +1}-1}{p_i-1} $$ similary $$\int \frac{n!}{j!(n-j)!}=\prod_{i\geq1} \frac{p_i^{\sum_{k\geq 1} \lfloor {n\over p_i^k}\rfloor-\lfloor {j\over p_i^k}\rfloor-\lfloor {n-j\over p_i^k}\rfloor+1}-1}{p_i-1}$$
