Solve the following equation, where $b$ and $n$ are natural numbers, and $b>n$.

$$
\frac{x-(b-1)}{1}+\frac{x-(b-2)}{2}+\frac{x-(b-3)}{3}+\ldots+\frac{x-(b-n)}{n}=\frac{x-1}{b-1}+\frac{x-2}{b-2}+\frac{x-3}{b-3}+\ldots+\frac{x-n}{b-n}
$$
$$
\frac{x-(b-1)}{1}+\frac{x-(b-2)}{2}+\frac{x-(b-3)}{3}+\ldots+\frac{x-(b-n)}{n}=\frac{x-1}{b-1}+\frac{x-2}{b-2}+\frac{x-3}{b-3}+\ldots+\frac{x-n}{b-n}
$$

Due to the condition $b>n$, the denominators on the right side of (1) are never zero. There are $n$ terms on both sides, and if we subtract 1 from each, (1) transforms into:

$$
\frac{x-b}{1}+\frac{x-b}{2}-\ldots+\frac{x-b}{n}=\frac{x-b}{b-1}+\frac{x-b}{b-2}+\ldots+\frac{x-b}{b-n}
$$

From this expression, it is clear that $x=b$ is always a solution to (1). Furthermore, if

$$
\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n} \neq \frac{1}{b-1}+\frac{1}{b-2}+\ldots+\frac{1}{b-n}
$$

then (1) has no other solutions besides this. However, if equality holds in (3), then (2) and thus (1) hold for all real $x$.

Let's try to simplify the condition (3)! Due to the condition $b>n$, we have $1/(b-n) \leq 1/1, \ldots, 1/(b-1) \leq 1/n$, so the right side of (3) is at most as large as the left side, and equality holds only if $b-n=1$, i.e., $b=n+1$. Thus, (3) is equivalent to the condition

$$
b \neq n+1
$$

Therefore, if $b \neq n+1$, then (1) has only one solution $x=b$; if $b=n+1$, then (1) is satisfied by all real numbers.