Special prime of odd perfect number Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e.,$q$  is prime with $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$ . (That is, $\sigma(N)=2N$ where $\sigma$ is the classical sum-of-divisors function.)
I can easily prove that $q$ not the smallest divisor of $N$ :  

Proof with contradiction Assume that $q$ be the smallest divisor of $N$ then because $k$ is an odd number. So $(q+1)$ divides $\sigma(q^k)$ and then $(q+1)$ has a factor less than $q$ 

Is it possible that $q$ be the largest Divisor of $N$ ?
 A: Not a complete answer, just some remarks that would be too long to fit in the Comments section.
This is a good question to ask.
The assertion that the Euler or special prime $q$ of an odd perfect number $N=q^k n^2$ is also its largest prime factor is a folklore conjecture in the theory of OPNs, together with the conjecture that $k=1$, which was conjectured by Descartes, Frenicle, and subsequently by Sorli.  Both conjectures would have followed from a proof of the inequality $n < q$ (see [Dris, 2012]).  Alas, in two (separate and) recent preprints, [Brown, 2016] and [Starni, 2017] claim a proof for the inequality $q < n$.
Here is some data supporting both $q < n$ and $k = 1$:
OEIS sequence A228059:  Odd numbers of the form $p^{1+4s}{r^2}$, where $p$ is prime of the form $1+4m$, $r > 1$, and $\gcd(p,r) = 1$ that are closer to being perfect than previous terms.
Here are the first couple of terms:
$$45 = 5\cdot{3^2}$$
$$405 = 5\cdot{3^4}$$
$$2205 = 5\cdot(3\cdot7)^2$$
$$26325 = 13\cdot({3^2}\cdot5)^2$$
$$236925 = 13\cdot({3^3}\cdot5)^2$$
$$1380825 = 17\cdot(3\cdot5\cdot19)^2$$
$$1660725 = 61\cdot(3\cdot5\cdot11)^2$$
$$35698725 = 61\cdot({3^2}\cdot5\cdot17)^2$$
$$3138290325 = 53\cdot({3^4}\cdot5\cdot19)^2$$
Notice that, except for the first term, all terms (so far) satisfy $p = q < n = r$, and all such $p$'s have exponent $1$ (i.e. $s=0$).  Note further that some of the terms have $p = q$ as the largest prime factor while others do not.  Lastly, notice that, except for the last term, all terms (so far) satisfy $r = n < q^2 = p^2$.
So to answer your question:  It can happen that the Euler/special prime $q$ is the largest prime factor of $N$ (subject to some constraints), but we currently do not know of any proof.  That is, the conjecture that the Euler/special prime is the largest divisor of an odd perfect number may or may not be true (but not both, of course).
If you are interested in doing further readings on this topic, I invite you to check out the paper titled Euclid-Euler Heuristics for Perfect Numbers (and the references therein).
