Proving this quotient is always even.

Question

Let $n$ be an odd, composite integer.

Let $D$ equals the sum of the proper divisors of $n$ minus the last divisor, $d_j$. Also, let $n$ be a number such that $D \gt \frac{1}{2} \times{n}$. Let $d_i$ be the $i$th divisor of $n$. Also, let $j$ be the number of proper divisors and $j$ is odd.

So, $$D = 1 + d_1 + d_2 + ... + d_{j-1}$$

I want to prove that $$\frac{D}{d_1-1}$$ is always even.

Some facts I've figured out: $d_1$ is always prime; $d_i$ is always odd; $D$ is even since the sum of an even number of integers is even; and $d_1-1 \ge 2$ and is always even.

Edit: fixed wording

Answer 1

This makes no sense. $n = 125$ is odd and composite, $j=3$ is composite, the divisors are $1$, $5$, $25$, $D=6$ and $d_1 - 1$ = 4. $\dfrac D {d_1-1}$ is not an integer, much less an even integer.

Answer 2

This is false, the first number where $\frac{D}{d_1-1}$ is odd is $2,205$.