Find positive integers $a$ and $b$, with $a>b$, such that:

$$
\frac{2}{7}=\frac{1}{a}+\frac{1}{b}
$$
The equation is equivalent to $2 a b=7(a+b)$. Since 2 and 7 are coprime, it follows that $a b$ is a multiple of 7 and that $a+b$ is a multiple of 2. However, for $a b$ to be a multiple of 7, the only possibility is that $a$ is a multiple of 7 or $b$ is a multiple of 7. Let's first assume that $a$ is a positive multiple of 7, that is, $a=7 k$ for some positive integer $k$. From this, we obtain:

$$
2 a b=7(a+b) \Rightarrow 2 k b=7 k+b \Rightarrow(2 k-1) b=7 k
$$

This last equation implies that $b$ or $2 k-1$ are multiples of 7. If $b$ is a multiple of 7, $b=7 m$, and thus

$$
\frac{2}{7}=\frac{1}{a}+\frac{1}{b} \Rightarrow \frac{2}{7}=\frac{1}{7 k}+\frac{1}{7 m} \Rightarrow \frac{1}{k}+\frac{1}{m}=2
$$

However, $\frac{1}{k} \leq 1$ and $\frac{1}{m} \leq 1$. Thus $\frac{1}{k}+\frac{1}{m} \leq 2$. Therefore, the equation $\frac{1}{k}+\frac{1}{m}=2$ has a unique positive integer solution, namely, $k=1=m$. However, this solution does not interest us, as in this case $a=b$.

Let's then consider the case where $2 k-1$ is a multiple of 7.

One possibility for $2 k-1$ to be a multiple of 7 occurs when $k=4$. In this case, we have that

$$
\begin{gathered}
k=4 \Rightarrow a=7 k=28 \\
(2 k-1) b=7 k \Rightarrow 7 b=28 \Rightarrow b=4
\end{gathered}
$$

We then obtain

$$
\frac{2}{7}=\frac{1}{28}+\frac{1}{4}
$$