What are the constraints? For a string of size $$$N$$$ up to $$$10^6$$$ would pass with an easy upperbound:

The approximate number of divisors of a number N is $$$\sqrt[3]{N}$$$. For $$$N=10^6$$$, the value would be up to $$$100$$$ divisors only. Approximately half of the divisors is smaller than $$$\sqrt{N}$$$ and half is bigger than $$$\sqrt{N}$$$. That means, an easy upperbound with no knowledge needed in maths is:

$$$\frac{\sqrt[3]{N}}{2} \cdot \sqrt{N} + \frac{\sqrt[3]{N}}{2} \cdot N$$$ upperbound.

Do not forget that in real case scenarios, it would be lower than that. Substituting in the above upperbound would yield around $$$5 \cdot 10^7$$$ which is pretty good.

This sequence is the numbers $$$a_1, a_2, \ldots$$$ where $$$a_k$$$ is the smallest positive integer such that its sum of factors is more than $$$k$$$ times itself. You can see that $$$a_6$$$ is about $$$10^8$$$, so for practitcal constraints ($$$\leq 10^6$$$ probably) the sum of divisors of $$$n$$$ will be less than $$$6n$$$.

If x*y=n, you can write x as n/y. Now for every x, y is unique and less than or equal to n. So a very loose upper bound of sum of all x is nlogn.

In short, it's $$$O(n \log{\log{n}})$$$. Here's a sketch of a rough estimation:

The sum-of-divisors function $$$\sigma$$$ is a multiplicative function, so that for any positive integer $$$n$$$ with prime factorization $$$n = \prod_{i=1}^k p_i^{s_i}$$$, we have $$$\sigma(n) = \prod_{i=1}^k \sigma(p_i^{s_i}) = n \cdot \prod_{i=1}^k \frac{\sigma(p_i^{s_i})}{p_i^{s_i}}$$$. The sums-of-divisors for prime powers are easily calculated and seen to satisfy the inequality $$$\frac{p_i + 1}{p_i} \leq \frac{\sigma{p_i^{s_i}}}{p_i^{s_i}} \lt \frac{p_i}{p_i - 1}$$$. Taking logarithms and remembering a few things about convergent series from Calculus 2, we get that for some appropriate constant $$$C \in \mathbb{R}_{>0}$$$ which does not depend on $$$n$$$,

To get an upper bound, we can now make the rough estimation that the number $$$k$$$ of distinct prime factors of $$$n$$$ is at most $$$\log_2{n}$$$. This is a loose bound, but that's OK because it will be passed through a very slow-growing function: It in known that the sum of the reciprocals of the first $$$k$$$ primes is approximately $$$\ln (\ln{k}) + O(1)$$$. (For example, combine the Meissel-Mertens theorem with the prime number theorem. It's also clear that no set of $$$k$$$ distinct primes can do better.) So, we get that

for some appropriately chosen constant $$$D$$$ and all sufficiently large $$$n$$$, which is exactly the statement that $$$\sigma(n) \in O(n \log{\log{n}})$$$. To see that this bound isn't far from optimal (more precisely, to see $$$\sigma(n)$$$ is not in little-o $$$o(n \log{\log{n}})$$$), it suffices to consider values of $$$n$$$ that are large primorials and perform an identical computation with the left-hand side of the big inequality above.

UPDATE: The solution got accepted for n= 2*(10^5) with time taken 207 ms. Thanks everyone. 120975508