I miss some cases thanks for correcting @below

So are you saying that there are no solutions? Quite wrong since there are a few... Let $f(n)=\frac{n}{2^n}$ so that $f(1)=f(2)=\frac{1}{2}, f(3)=\frac{3}{8}$ etc. and we want to solve $f(a)=f(b)+f(c)$. Easy to see that $f$ is strictly decreasing for $n \ge 2$. Hence $b,c>a$. W.l.o.g. $c \ge b>a \ge 2$ (on recalling that $f(1)=f(2)$). If $b=c=a+1$, then $\frac{a}{2^a}=\frac{a+1}{2^{a+1}}$, contradiction! If $b,c \ge a+2$, then $\frac{a}{2^a} \le \frac{a+2}{2^{a+1}}$ and hence $a=2$ and $b=c=4$. We thus get the solutions $\boxed{(1,4,4)}, \boxed{(2,4,4)}$ and permutations. Finally, if $b=a+1, c \ge a+2$, then $\frac{a}{2^a} \le \frac{a+1}{2^{a+1}}+\frac{a+2}{2^{a+2}}$ and we get $a \le 4$. If $a=2, b=3$ or $a=3, b=4$, we must have $\frac{c}{2^c}=\frac{1}{8}$ absurd. Hence $a=4, b=5$ and we must have $\frac{c}{2^c}=\frac{3}{32}$ and hence $c=6$. We thus get the solution $\boxed{(4,5,6)}$ and permutations.

Sorry for disturbing. I just did 2023 CGMO today and found a more than annoying method to solve Problem 1. I'll only show the situation when b=c below because of the complexity.