How many (not-necessarily-prime) factors of $200!$ are there? First, get all the prime factors. (The maximum such factor will be less than $200$, of course--in fact it is $199$.) FactorInteger[200!]
gives each prime factor with its multiplicity:
{{2, 197}, {3, 97}, {5, 49}, {7, 32}, {11, 19}, {13, 16}, {17, 11}, {19, 10}, {23, 8}, {29, 6}, {31, 6}, {37, 5}, {41, 4}, {43, 4}, {47, 4}, {53, 3}, {59, 3}, {61, 3}, {67, 2}, {71, 2}, {73, 2}, {79, 2}, {83, 2}, {89, 2}, {97, 2}, {101, 1}, {103, 1}, {107, 1}, {109, 1}, {113, 1}, {127, 1}, {131, 1}, {137, 1}, {139, 1}, {149, 1}, {151, 1}, {157, 1}, {163, 1}, {167, 1}, {173, 1}, {179, 1}, {181, 1}, {191, 1}, {193, 1}, {197, 1}, {199, 1}}
So any factor can be expressed as a product $2^{a_2} \times 3^{a_3} \times \cdots \times 199^{a_{199}}$, where the value of $a_2$ can be $0, 1, 2, \ldots, 197$, and thus there are $198$ possible values in the $2$ "slot," and likewise $98$ possible values in the $3$ "slot," and so on. The total number of factors is then just the product of the total number of values for each "slot." That is computed thus:
Times @@ (# + 1 & /@ FactorInteger[200!][[All, 2]])
$n = 139503973313460993785856000000 = 1.39504\times 10^{29}$.
If you want the humongous list of factors, just compute $2^{a_2} \times 3^{a_3} \times \cdots$ and plug in every one of the $n$ combinations of values of $a_i$ given the constraints above.
Clearly this is not the way to solve Euler 608, but it does (I submit) solve the question as posed. Moreover, this method can be easily parallelized (by apportioning different ranges of $a_i$ to different processors), but this number is so large such direct computation is infeasible on realistic hardware.
Total@Table[ Sum[DivisorSigma[0, i k], {k, 1, 10^6}], {i, Divisors[4!]}]
. $\endgroup$