1. **Problem statement:** We need to find two consecutive natural numbers $n$ and $n+1$ such that the fractions $\frac{n}{7}$ and $\frac{n+1}{7}$ have the fraction $\frac{9}{11}$ lying between them on the number line. 2. **Understanding the problem:** Since $\frac{9}{11}$ lies between $\frac{n}{7}$ and $\frac{n+1}{7}$, it means: $$\frac{n}{7} < \frac{9}{11} < \frac{n+1}{7}$$ 3. **Find $n$:** Multiply all parts by 77 (the least common multiple of 7 and 11) to clear denominators: $$n \times 11 < 9 \times 7 < (n+1) \times 11$$ $$11n < 63 < 11n + 11$$ 4. **Solve inequalities:** - From $11n < 63$, we get $n < \frac{63}{11} = 5.727...$ - From $63 < 11n + 11$, we get $63 - 11 < 11n$, so $52 < 11n$, thus $n > \frac{52}{11} = 4.727...$ 5. **Determine $n$:** Since $n$ is a natural number, $n$ must satisfy: $$4.727... < n < 5.727...$$ So $n = 5$. 6. **Find the sum of the numerators:** The two numerators are $5$ and $6$, so their sum is: $$5 + 6 = 11$$ **Final answer:** 11