5. (10 points) If $\frac{O}{11}-\frac{7}{\square}<\frac{4}{5}$ holds, then the maximum sum of the non-zero natural numbers that can be filled in “ $O$ ” and “ $\square$ ” is


393
【Analysis】Transform the inequality $\frac{\bigcirc}{11}-\frac{7}{\square}<\frac{4}{5}$ by finding a common denominator, which becomes $55 \times \square$, i.e., $\frac{5 \times \bigcirc \times \square-385}{55 \times \square}<\frac{44 \times \square}{55 \times \square}$, then $5 \times \bigcirc \times \square-385<44 \times \square$, i.e., $5 \times \bigcirc \times \square-44 \times \square<385$, $\square \times(5 \times \bigcirc-44)<385$, when $5 \times \bigcirc$ is close to 44, the value of $\square$ is larger, so when $\bigcirc=9$, the value of $\square$ is the largest; solve accordingly.
【Solution】Solution: Transform the inequality $\frac{\bigcirc}{11}-\frac{7}{\square}<\frac{4}{5}$ by finding a common denominator,
i.e., $\frac{5 \times \bigcirc \times \square-385}{55 \times \square}<\frac{44 \times \square}{55 \times \square}$,
then, $5 \times \bigcirc \times \square-385<44 \times \square$,
i.e., $5 \times \bigcirc \times \square-44 \times \square<385$,
$\square \times(5 \times \bigcirc-44)<385$,
when $5 \times \bigcirc$ is close to 44, the value of $\square$ is larger,
so when $\bigcirc=9$, the value of $\square$ is the largest,
$\square \times(5 \times 9-44)<385$
$<385$
the maximum is 384,

Therefore, when $\bigcirc=9, \square=384$, the sum of the two numbers is the largest,
i.e., $\bigcirc+\square=9+384=393$.
Answer: The maximum sum of the non-zero natural numbers that can be filled in “$\bigcirc$” and “$\square$” is 393.
The answer is: 393.