Let $a$ and the larger $b$ be positive integers. Calculate the sum of the irreducible fractions with a denominator of 7 that lie between $a$ and $b$.
I. solution: The sequence of fractions with a denominator of 7 (including both $a$ and $b$)

$$
\frac{7 a}{7}, \frac{7 a+1}{7}, \frac{7 a+2}{7}, \ldots, \frac{7 b-1}{7}, \frac{7 b}{7}
$$

However, among the terms of this sequence,

$$
\frac{7 a}{7}, \frac{7 a+7}{7}, \ldots, \frac{7 b}{7}
$$

are reducible fractions, and can be written in the following form

$$
a, a+1, \ldots, b
$$

The sum of the irreducible fractions with a denominator of 7 between $a$ and $b$ is given by the difference between the sums of sequences (1) and (2).

The sequence (1) is an arithmetic sequence whose first term is $a$, common difference is $\frac{1}{7}$, and last term is $b$. From these data, the number of terms can be obtained using the formula for the $n$-th term:

$$
b=a+(n-1) \frac{1}{7}, \quad \text { hence } \quad n=7 b-7 a+1
$$

Thus, the sum of the sequence (1) is, according to the sum formula:

$$
S'=\frac{(7 b-7 a+1)(b+a)}{2}
$$

The sequence (2) is the sequence of natural numbers starting from $a$ and ending at $b$, so its sum is directly given by

$$
s=\frac{(b-a+1)(b+a)}{2}
$$

Therefore, the sum of the fractions with the desired property is

$$
S=S'-s=\frac{(b+a)(7 b-7 a+1-b+a-1)}{2}=\frac{(b+a)(6 b-6 a)}{2}=3\left(b^{2}-a^{2}\right)
$$

II. solution: The sum of the 6 fractions with a denominator of 7 between $a$ and $a+1$

$$
\left(a+\frac{1}{7}\right)+\left(a+\frac{2}{7}\right)+\ldots+\left(a+\frac{6}{7}\right)=6 a+\frac{21}{7}=6 a+3
$$

In each subsequent interval, the fractions increase by one unit, so the sum of the 6 fractions increases by 6.

Therefore, the sum of the irreducible fractions with a denominator of 7 between $a$ and $b$ is the sum of an arithmetic sequence whose first term is $6 a+3$, common difference is 6, and number of terms is $b-a$.

Thus, according to the sum formula, we get

$$
\begin{gathered}
S=\frac{b-a}{2}[2(6 a+3)+(b-a-1) 6]=(b-a)(6 a+3+3 b-3 a-3)= \\
=(b-a)(3 a+3 b)=3\left(b^{2}-a^{2}\right)
\end{gathered}
$$

III. solution: The non-reducible fractions involved do not form an arithmetic sequence, but their sum can first be written in increasing order of terms, and then in decreasing order:

$$
\begin{aligned}
& S=\left(a+\frac{1}{7}\right)+\left(a+\frac{2}{7}\right)+\ldots+\left(a+\frac{6}{7}\right)+\left(a+\frac{8}{7}\right)+\ldots+\left(b-\frac{2}{7}\right)+\left(b-\frac{1}{7}\right) \\
& S=\left(b-\frac{1}{7}\right)+\left(b-\frac{2}{7}\right)+\ldots+\left(b-\frac{6}{7}\right)+\left(b-\frac{8}{7}\right)+\ldots+\left(a+\frac{2}{7}\right)+\left(a+\frac{1}{7}\right)
\end{aligned}
$$

The sum of the terms that are directly below each other is always $b+a$, and there are 6 such pairs of terms between two consecutive integers. Thus, we get

$$
2 S=(b-a) \cdot 6 \cdot(b+a)=6\left(b^{2}-a^{2}\right), \quad \text { or } \quad S=3\left(b^{2}-a^{2}\right)
$$