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 $R\equiv r(\mathrm{mod}9)$, $r\in [1,9]$. Then $N\equiv 2345r\equiv 5r(\mathrm{mod}9)$ 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 $N\equiv 5678s\equiv 8s(\mathrm{mod}9)$. So $5r\equiv N\equiv 8s(\mathrm{mod}9)$. Letting $r$ run through $1,2,\dots ,9$ 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.