National Olympiad of Argentina

Such numbers exist. The least one is $11725$.
Let $N$ be the sum of $2345$ positive integers with digit sum $R$, and let $R \equiv r \pmod{9}$, $r \in [1, 9]$. Then $N \equiv 2345r \equiv 5r \pmod{9}$ as each summand is congruent to $r$ modulo $9$. Similarly if $N$ is the sum of $5678$ numbers with digit sum congruent to $s$ modulo $9$, $s \in [1, 9]$, then $N \equiv 5678s \equiv 8s \pmod{9}$. So $5r \equiv N \equiv 8s \pmod{9}$. Letting $r$ run through $1, 2, \ldots, 9$ yields the admissible pairs of remainders: $(1, 4)$, $(2, 8)$, $(3, 3)$, $(4, 7)$, $(5, 2)$, $(6, 6)$, $(7, 1)$, $(8, 5)$, $(9, 9)$.

Note that $N \ge \max(2345r, 5678s)$ for every such pair $(r, s)$ because the least number with digit sum congruent to $r$ or $s$ modulo $9$ is $r$ or $s$ respectively. If $r = 5$, $s = 2$ the last observation gives $N \ge \max(2345 \cdot 5, 5678 \cdot 2) = 11725$. For each remaining pair $(r, s)$ one of the numbers $2345r$ and $5678s$ is greater than $11725$. It follows that the least $N$ in question, if it exists, is at least $11725$.
On the other hand $N = 11725$ is possible. Indeed $11725 = 2345 \cdot 5$ is equal to the sum of $2345$ numbers equal to $5$. Also $11725$ is equal to the sum of $5678$ numbers with digit sum $2$: $41$ summands $11$ and $5637$ summands $2$ ($41 \cdot 11 + 5637 \cdot 2 = 11725$). In all there are $41 + 5637 = 5678$ summands with digit sum $2$, as needed.