(given to Charlotte Le Mouel). The integers $a$ and $b$ are coprime, with $a > b$. Compare the numbers:

$$
m=\left\lfloor\frac{a}{b}\right\rfloor+\left\lfloor\frac{2 a}{b}\right\rfloor+\cdots+\left\lfloor\frac{(b-1) a}{b}\right\rfloor
$$

and:

$$
n=\left\lfloor\frac{b}{a}\right\rfloor+\left\lfloor\frac{2 b}{a}\right\rfloor+\cdots+\left\lfloor\frac{(a-1) b}{a}\right\rfloor
$$

where $\lfloor x\rfloor$ denotes the integer part of the real number $x$.
. Method 1. $\lfloor k a / b\rfloor$ is the greatest integer less than or equal to $k a / b$, so $\frac{k a}{b}-1<\left\lfloor\frac{k a}{b}\right\rfloor \leqslant \frac{k a}{b}$. This can be written as $\left\lfloor\frac{k a}{b}\right\rfloor=\frac{k a}{b}-\frac{u_{k}}{b}$, where $u_{k}$ is the integer between 0 and $b-1$ such that $k a-u_{k}$ is divisible by $b$ (the remainder of the Euclidean division). The sum of the $\frac{k a}{b}$ is:

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

But we need to subtract the sum of the $\frac{u_{k}}{b}$: it is not possible to calculate $u_{k}$ for a given $k$, but we can assert that the $u_{k}$ are non-zero and do not take the same value twice. Indeed, if $u_{k}=u_{k^{\prime}}$, this would mean that $k a-u_{k}$ and $k^{\prime} a-u_{k}$ are both divisible by $b$, so $(k-k^{\prime}) a$ is divisible by $b$: since $b$ is coprime with $a$, it would divide $k-k^{\prime}$ (by Gauss's theorem), which is not possible if $k$ and $k^{\prime}$ are distinct and both between 0 and $b-1$ (so $0<\left|k-k^{\prime}\right|<b$). The same reasoning proves that $u_{k}$ cannot be zero. And if $u_{k}$ takes $b-1$ distinct values between 1 and $b-1$, it takes all the values from 1 to $b-1$, so the sum of the $\frac{u_{k}}{b}$ is:

$$
1+2+\cdots+\frac{b-1}{b}=\frac{b-1}{2}
$$

The sum $m$ is therefore equal to $\frac{(a-1)(b-1)}{2}$. The same calculation on the sum $n$ will give the same result, symmetric in $a$ and $b$, so we will necessarily have $m=n=\frac{(a-1)(b-1)}{2}$. Method 2. We can also note that for all $k, \frac{k a}{b}+\frac{(b-k) a}{b}=a$ is an integer. If neither $\frac{k a}{b}$ nor $\frac{(b-k) a}{b}$ are integers, the sum of their decimal parts is 1, so the sum of their integer parts is $a-1=2 \frac{a-1}{2}$. If $b$ is even, $a$, coprime with $b$, is odd, and the middle term, $\left\lfloor\frac{a}{2}\right\rfloor$, is $\frac{a-1}{2}$. We deduce that the sum is not changed by replacing each integer part with $\frac{a-1}{2}$, and since there are $b-1$ integer parts, their sum is always $\frac{(a-1)(b-1)}{2}$, both in the calculation of $m$ and in the calculation of $n$.