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:

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

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 $452$, $453$, $\ldots$, $1809$ 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 $\{452, \ldots, 602\}$. In addition one of them is in $\{906, \ldots, 1207\}$ because the three largest numbers in $\{452, \ldots, 602\}$ have sum less than $1810$. If, e.g., $x \ge 906$ then $y+z \le 904$; in particular $y, z \in \{452, \ldots, 602\}$. 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.