What fraction of $\frac{4}{7}$ must be added to itself to make the sum $\frac{15}{14}$?

Let the required fraction of $\frac{4}{7}$ be $x$.
According to the given condition, this fraction of $\frac{4}{7}$ is added to $\frac{4}{7}$ itself to give a sum of $\frac{15}{14}$:

$$\frac{4}{7}+x\cdot \left(\frac{4}{7}\right)=\frac{15}{14}$$

Taking $\frac{4}{7}$ as a common factor out from the left-hand side:

$$\frac{4}{7}(1+x)=\frac{15}{14}$$

Multiplying both sides by $\frac{7}{4}$:

$$1+x=\frac{15}{14}\times \frac{7}{4}$$

$$1+x=\frac{15}{2\times 4}$$

$$1+x=\frac{15}{8}$$

Subtracting $1$ from both sides to solve for $x$:

$$x=\frac{15}{8}-1$$

$$x=\frac{15-8}{8}=\frac{7}{8}$$

Thus, the fraction that must be multiplied and added is $\frac{7}{8}$.