I've just been able to prove that any odd fraction can be represented as the finite sum of different odd unit fractions.
First, note that it is sufficient to prove the following $(\star)$ :
$(\star)\ \ $ Any natural number $N$ can be represented as the finite sum of different odd unit fractions.
(If we get $(\star)$, then all we need is to divide $N$ by an odd which is the very denominator.)
Now, let $p_i$ be the $i$-th odd prime number arranged in ascending order. ($p_1=3, p_2=5, p_3=7, \cdots$) Here, we use the following two famous facts :
$(1) \ \ p_{i+1}\lt 2p_i$ (by Chebyshev)
$(2) \ \frac1{p_1}+\frac1{p_2}+\cdots$ diverges. (by Euler)
By $(2)$, we can take $n\ge 5$ such that
$$\frac1{p_1}+\frac1{p_2}+\cdots+\frac1{p_n}\gt N+1.$$
Let's fix one of such $n$. Then, letting $P=p_1p_2\cdots p_n$, we know that $P\gt 3$ and that
$$\left(1+\frac1{p_1}\right)\cdots\left(1+\frac1{p_n}\right)\gt 1+\left(\frac1{p_1}+\cdots+\frac1{p_n}\right)\gt 1+(N+1)\gt N+1+\frac 3P.$$
Now, let $q_j$ be the $j$-th positive divisor of $P$ arranged in ascending order: ($q_1, q_2,\cdots,q_m$ where $m=2^n$)
$$q_1=1, q_2=3,q_3=5,q_4=7, q_5=11, q_6=13, q_7=15, \cdots, q_n=P.$$
Lemma 1 : $q_{i+1}\lt 2q_i$ when $i=2,\cdots,m-2$.
(Note that this lacks $i=1,m-1$ because $\frac{q_2}{q_1}=\frac{q_m}{q_{m-1}}=3\gt 2$.)
Lemma 2 : Each of $3,4,\cdots, q_1+q_3+q_4+\cdots+q_i$ can be represented as the sum of some different elements in $q_1,q_2,\cdots,q_i$ when $i=3,\cdots,m-1$.
(For example, when $i=3$, we get $3=q_2, 4=q_1+q_2, 5=q_3, 6=q_1+q_3$. When $i=4$, we get $7=q_4, 8=q_1+q_4,9=q_1+q_2+q_3, 10=q_2+q_4, 11=q_1+q_2+q_4, 12=q_3+q_4, 13=q_1+q_3+q_4.)$
(I'll prove these lemmas at the last.)
Let's prove that $(\star)$ follows these lemmas. To do this, let $R=q_1+q_3+q_4+\cdots+q_{m-1}$.(Note that this lacks both of $q_2=3$ and $q_m=P$) Then, we get
$$R=(q_1+q_2+\cdots+q_m)-(3+P)=(p_1+1)\cdots(p_n+1)-(3+P).$$
Hence, since we get
$$\frac RP=\frac{(p_1+1)\cdots (p_n+1)}{P}-\left(\frac 3P+1\right)=\left(1+\frac1{p_1}\right)\cdots\left(1+\frac1{p_n}\right)-\left(\frac 3P+1\right)\gt\left(N+1+\frac 3P\right)-\left(\frac 3P+1\right)=N,$$
we know that $3\le NP\lt R$. By lemma 2, we can represent $NP$ as
$$NP=q_{a_1}+q_{a_2}+\cdots+q_{a_k} \ \ (1\le a_1\lt aa_2\lt \cdots\lt a_k\le m-1).$$
Hence, we get
$$N=\frac{q_{a_1}}{P}+\cdots+\frac{q_{a_k}}{P}.$$
Here, since each of $q_{a_i}$ is a positive divisor of $P$, each of $\frac{q_{a_i}}{P}$ is an odd unit fraction. Now the proof for $(\star)$ is completed.
In the following, we are going to prove the above lemmas.
Proof for lemma 1 : Let $_kr_i$ be the $i$-th number in ascending order which can be represented the product of $k\ (k=0,\cdots,n)$ different elements in $p_1,\cdots,p_n$.
For example, $$_kr_1=p_1p_2\cdots p_k,\ _kr_{\binom{n}{k}}=p_np_{n+1}\cdots p_{n-k+1},\ _0r_1=1, \ _nr_1=P.$$
Then, by the fact $(1)$, we know that we can get $_kr_{j+1}\lt 2 _kr_j$.
Now, let's prove that $q_{i+1}\lt 2q_i$ for $i=2,\cdots, m-2$. Note that $q_i=_kr_j$ for some $k,j$ where $1\le k\le n-1$. First, if $j\lt \binom nk$, then we know that
$$q_{i+1}\le _kr_{j+1}\lt 2_kr_j=2q_i.$$
Next, if $j=\binom nk$, then we get
$$q_i=p_n\times p_{n-1}\times \cdots\times p_{n-k+1}$$
where $i\le m-2$ leads $k\le n-2$ and $n-k+1\ge 3$. In the $n=5$ case, by $p_n=p_5=13$ and $p_1\times p_2=3\times 5=15$, we know
$$q_{i+1}\le p_1\times p_2\times p_{n-k+1}\times p_{n-k+2}\times\cdots\times p_{n-1}=\frac{15}{13}p_{n-k+1}\times p_{n-k+2}\cdots\times p_{n-1}\times p_n=\frac{15}{13}q_i\lt 2q_i.$$
In the $n\ge 6$ cases, by $p_n\ge p_6=17\gt 15=p_1\times p_2$, we get
$$_{k+1}r_1\lt _kr_{\binom nk}\lt _{k+1}r_{\binom{n}{k+1}}.$$
Hence, since we can tak $l$ such that
$$_{k+1}r_{l}\lt _kr_{\binom nk}\lt _{k+1}r_{l+1},$$
we know that
$$q_{i+1}\le _{k+1}r_{l+1}\lt 2 _{k+1}r_{l}\lt 2q_i$$
for $q_i= _kr_{\binom nk}.$ Now the proof for lemma 1 is completed.
Proof for lemma 2 : The $i=3$ case is obvious. We treat $l\ge 4$ in the following. First, by induction on $i$, we are going to prove
$$(\star\star)\ \ 3+q_{i+1}\le q_1+q_3+q_4+\cdots+q_i+1$$
when $4\le i\le m-2$.
In $i=4$ case, we get
$$LHS=3+q_5=3+11=14, RHS=q_1+q_3+q_4+1=1+5+7+1=14.$$
Now, supposing that $(\star\star)$ is true when $i=k\ (4\le k\le m-3)$, in $i=k+1$ case, by lemma 1 and inductive supposition, $LHS=3+q_{k+2}\lt 3+2q_{k+1}=(3+q_{k+1})+q_{k+1}\lt (q_1+q_3+q_4+\cdots+q_k+1)+q_{k+1}=RHS.$
Hence, the proof for $(\star\star)$ is completed.
Now, let's prove lemma 2 by induction on $i$. The $i=4$ case has been already proved, so let's suppose that lemma 2 is true when $i=k\ (4\le k\le m-2)$. We know that the numbers from $3+q_{k+1}$ to $q_1+q_3+q_4+\cdots+q_{k+1}$ can be represented as the sum of some different elements in $q_1,q_2, \cdots, q_{k+1}.$ By the way, by $(\star\star)$, since
$$3+q_{k+1}\le q_1+q_3+q_4+\cdots+q_k+1,$$
as a result we know that the numbers from $3$ to $q_1+q_3+q_4+\cdots+q_{k+1}$ can be represented as the sum of some different elements in $q_1,q_2,\cdots,q_{k+1}$. So lemma 2 is true when $i=k+1$. Hence we know that lemma 2 is true for $4\le i\le m-1$. Now the proof for lemma 2 is completed.
