Let $q(n)$ be the number of integers $m<n$ such that $m$ is square-free. Let $p(n)$ be the number of integers $m<n$ such that the sum of the prime factors of $m$ is square-free. And let $s(n)$ be the number of integers $m<n$ such that the sum of the divisors of $m$ is square-free. The table shows values of $q$, $p$ and $s$ for some $n$:
n q p s
100 60 68 24
1000 607 660 157
10000 6082 6343 1090
100000 60793 62352 8097
1000000 607925 618969 64306
It's known that the distribution is asymptotically equivalent with a straight line and that $\frac{6}{\pi^2}\cdot 100\approx 60.79$ percent of all integers are square-free. So how to explain this anomaly?
Why is the distribution of square-free sums of divisors so low?
Expanding on hardmath's insightful comment, another necessary condition for $\sigma(n)$ to be square-free is that the values of $\sigma(p^k)$ are pairwise relatively prime. For instance, almost all of the prime power factors of $n$ would need to have even exponents in order for $\sigma(n)$ to avoid being divisible by $4$. Consider also that the majority of $n$ have many prime divisors (about $\log \log n$ typically). So $n$ needs to be "almost" a perfect square (except for one allowable exceptional odd prime and the prime $2$).
This is really quite rare and it would be unreasonable to expect any positive proportion of $n$ to satisfy such constraints. It’s fairly easy to prove that the density of such $n$ is only $O(x/\log x)$ up to $x$, similar to that of the primes (but with a larger constant in front).