74th Romanian Mathematical Olympiad

a) Since $1+1+1+3+3=1\cdot 1\cdot 1\cdot 3\cdot 3$, there exist $5$ odd numbers whose sum is equal to their product, so $5$ is special.

b) If $n$ is a special number, then there exist the odd numbers $a_1,a_2,\dots ,a_n$ such that $a_1+a_2+\cdots +a_n=a_1{a}_2\dots a_n$. Let's assume that, among these, $k$ are of the form $M_4+3$ and the remaining $n-k$ are of the form $M_4+1$. Then $a_1+a_2+\cdots +a_n=M_4+3k+1\cdot (n-k)=M_4+2k+n$, (1). Since the product of two odd numbers has the form $M_4+1$ if the numbers leave the same remainder when divided by $4$, and the form $M_4+3$ otherwise, we infer that the product $a_1{a}_2\dots a_n$ has the form $M_4+1$ when $k$ is even, and the form $M_4+3$ when $k$ is odd, (2). Since $a_1+a_2+\cdots +a_n=a_1{a}_2\dots a_n$, (1) and (2) yield $n=M_4+1$.

If $n=4t+1$, $t\in {\mathbb{N}}^{\ast }$, for $a_1=a_2=\cdots =a_{n-2}=1$, $a_{n-1}=3$ and $a_n=2t+1$, we have $a_1+a_2+\cdots +a_n=a_1{a}_2\dots a_n=6t+3$, so any number of the form $M_4+1$ is special. This shows that the special numbers are those of the form $M_4+1$. The set $\{2,3,\dots ,2024\}$ contains $505$ numbers of the form $M_4+1$, so it contains $505$ special numbers.