Is that true : Every positive rational number $q$ can be written as $q = \sum_{i=0}^{k}1/n_i$ , where $n_i,k$ are positive intergers and $n_i\not=n_j$ if $i\not=j$.

Express the fraction $\dfrac{a}{b}$ as a sum of $a$ copies of $\dfrac{1}{b}$. (For technical reasons, if $b=1$, use $2a$ copies of $\dfrac{1}{2}$.) Major flaw: the denominators are not all different.
Using repeatedly the identity $$\frac{1}{k}=\frac{1}{k+1}+\frac{1}{k(k+1)},\tag{1}$$ we can express any $\dfrac{1}{n}$ as a sum of distinct unit fractions with all denominators as large as we wish.
So leave the first $\dfrac{1}{b}$ alone. Express the second one as $\dfrac{1}{b+1}+\dfrac{1}{b(b+1}$. For the third $\dfrac{1}{b}$, use Identity $(1)$ repeatedly to express $\dfrac{1}{b}$ as a sum of distinct unit fractions with denominators all greater than $b(b+1)$. Continue.