Argentina_2018

Such a partition clearly needs at least $2$ regular sets (e.g., $900$ and $910$ must be in different regular sets). Here is a partition with $2$ regular sets:

$$A=\{452,\dots ,602\}\cup \{906,\dots ,1207\},B=\{603,\dots ,905\}\cup \{1208,\dots ,1809\}.$$

Let us check that $A$ and $B$ are indeed regular. Suppose on the contrary that one of them has a subset $X$ with sum $1810$. Note that $X$ has at most $3$ elements as the sum of the $4$ smallest numbers among $452$, $453$, $\dots$, $1809$ is $452+453+454+455>1810$. Let $X$ have exactly $3$ elements $x,y,z$ with $x+y+z=1810$. If $X\subset B$ then $x+y+z\ge 603+604+605>1810$, which is impossible. Hence $X\subset A$ and clearly one of $x,y,z$ is in $\{452,\dots ,602\}$. In addition one of them is in $\{906,\dots ,1207\}$ because the three largest numbers in $\{452,\dots ,602\}$ have sum less than $1810$. If, e.g., $x\ge 906$ then $y+z\le 904$; in particular $y,z\in \{452,\dots ,602\}$. However then $y+z\ge 452+453=905$, a contradiction.

Let $X$ have exactly $2$ elements $x,y$ with $x+y=1810$ and $x<y$; then $x\le 904$, $y\ge 906$. It follows that if $X\subset A$ then $x\le 602$. On the other hand $y\le 1207$, so that $x+y\le 602+1207<1810$. Similarly if $X\subset B$ then $y\ge 1208$. Because $x\ge 603$, this yields $x+y\ge 603+1208>1810$. In both cases we reach a contradiction.

Because $X$ has more than $1$ element, the conclusion is that $A$ and $B$ are both regular.