For all integers $n \geqslant 0$, we denote $s(n)$ as the sum of the digits of $n$. Determine all integers $n \geqslant 0$ such that $n \leqslant 2 s(n)$.
Let $n$ be a natural number and let $a_{d} a_{d-1} \ldots a_{1} a_{0}$ be its decimal representation, meaning that $a_{0}$ is the units digit, $a_{1}$ is the tens digit, etc...

First, if $n$ has only one digit, then $n=a_{0} \leqslant 2 a_{0}$, so $n$ is a solution to the problem, and the numbers $0,1,2, \ldots, 9$ are solutions.

We now assume that $n$ has at least two digits.

Then $n=a_{d} \cdot 10^{d}+a_{d-1} \cdot 10^{d-1}+\ldots a_{1} \cdot 10^{1}+a_{0} \cdot 10^{0}$ and $s(n)=a_{d}+\ldots+a_{1}+a_{0}$. If $d \geqslant 2$, then for all $1 \leqslant i \leqslant d-1, a_{i} \cdot 10^{i} \geqslant a_{i} \cdot 2$ and since $a_{d}$ is non-zero, $a_{d} \cdot 10^{d}=2 a_{d}+\left(10^{d}-2\right) \cdot a_{d} \geqslant 2 a_{d}+10^{d}-2 > 2 a_{d}+a_{0}$ because $a_{0} < 10$. Therefore,

$$
n = a_{d} \cdot 10^{d} + a_{d-1} \cdot 10^{d-1} + \ldots + a_{1} \cdot 10^{1} + a_{0} \cdot 10^{0} > 2 a_{d} + a_{0} + 2 a_{d-1} + \ldots + 2 a_{1} + a_{0} = 2 s(n)
$$

Thus, if $n$ has three or more digits, $n$ is not a solution.

We assume that $n$ has two digits. If $a_{1} \geqslant 2$, then

$$
n=a_{1} \cdot 10+a_{0}=2 a_{1}+a_{0}+8 \cdot a_{1} \geqslant 2 a_{1}+a_{0}+8 \cdot 2 > 2 a_{1}+a_{0}+a_{0}=2 s(n)
$$

so $n$ is not a solution to the problem. We deduce that $a_{1}=1$. We can then easily test all the numbers between 10 and 19 to see which ones satisfy the problem. Alternatively, we can note that the inequality $n \leqslant 2 s(n)$ can be rewritten as $10 \cdot 1+a_{0} \leqslant 2\left(1+a_{0}\right)$, which simplifies to $a_{0} \geqslant 8$. Therefore, the only two-digit numbers that satisfy the problem are 18 and 19. Conversely, we have $18 \leqslant 2(1+8)$ and $19 \leqslant 2(1+9)$.

The solutions sought are therefore $\{0,1,2,3,4,5,6,7,8,9,18,19\}$.

Comment: The problem is well understood, students quickly have an intuition of the result and the reasoning to be made, but it is more difficult to implement it rigorously. The upper bound of $s(n)$ was a good idea. It is important to avoid introducing results, even correct ones, when they are not immediate: in this case, proving that for all $k \geqslant 3$, $18 k < 10^{k-1}$ was one of the main difficulties of the problem, so a proof of this result was expected. Finally, many students forget that 0 is a solution.