Preamble - This question is an offshoot from the following earlier questions here at MSE:
Can an odd perfect number be divisible by 825?
Can an odd perfect number be divisible by 165?
My own question is the following:
Can an odd perfect number be divisible by $101$?
Here is my own (quick) attempt at a (partial) answer:
Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e. $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$). Also, let $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$ be the canonical (i.e., prime) factorization of $N$.
If $q = 101$, then since $$(q+1)\mid\sigma(q^k) \mid \sigma(N)=2N,$$ then we have $$3\cdot17 = 51 = \frac{q+1}{2} \mid N.$$
Thus, it follows that $$3 \mid N \Longrightarrow 3^2 \mid N,$$ and $$17 \mid N \Longrightarrow {17}^2 \mid N.$$
Now test: (Note that $\alpha_i \geq \beta_i, \forall i$.) $$2 = I(N) \geq \prod_{i=1}^{\omega(N)}{\left(1 + \frac{1}{p_i} + \ldots + \frac{1}{{p_i}^{\beta_i}}\right)} \geq \frac{102}{101}\cdot\frac{13}{9}\cdot\frac{307}{289} = \frac{407082}{262701} \approx 1.549602019.$$
This is as far as I could go.
Of course, even after establishing $q \neq 101$, we still have to consider the remaining case $101 \mid n \mid N$, whence it follows that ${101}^2 \mid n^2 \mid N$.