Suppose you have some positive, odd, composite integer $n$. Let $$ D = d_1 + d_2 +d_3+...+d_{j-1} $$
where $d_i$ is some divisor of $n$ and $j$ is an odd number corresponding to the number of proper divisors of $n$. $d_j$ is the last proper divisor of $n$. Assuming that $D \gt \frac{n}{2}$ holds, is there some way to prove that $$ \frac{D}{d_2-1} \neq d_j $$
for any $n$ that satisfies the aformentioned constraints? I know that $D$ is always even, $d_2$ is always prime and is at least $3$. I am only concerned about when the quotient is odd which seems to be much rarer than when it is even.
For example: if $n=2205$, then $D=1506$ and $d_2-1=2$. $$\frac{D}{d_2-1}=753\neq d_j = 735$$
This is not really an answer, just some comments that are too long to fit in the appropriate section. In what follows, I shall be assuming that $$d_1 < d_2 < d_3 < \ldots < d_{j-1} < d_j$$ holds.
Suppose that $n$ is an odd perfect number. Euler proved that a hypothetical odd perfect number must have the form $$n = p^k m^2$$ where $p$ is the special/Eulerian prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
So, as you can see, your problem essentially reduces to determining factors of $m$, which is known to be composite (in fact, Nielsen has shown that $\omega(m) \geq 9$). Alas, even with modern computer technology and innovations in factor/sigma-chain methods, no factor chain has terminated that produces a proof for divisibility by a (certain) prime (say, less than $105$).
Alternatively, let $s(n)=\sigma(n)-n$ denote the aliquot sum (or sum of proper divisors) of $n$. As before, $n$ has the form $n = p^k m^2$. Then we obtain $$D + d_j = s(n) = n$$ $$D + d_j = \sigma(p^k)\sigma(m^2) - p^k m^2,$$ where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.
Once more, you see the paramount importance of determining factors of the component $m$.