1. Let the sum of the digits of the natural number $x$ be $S(x)$, then the solution set of the equation $x+S(x)+S(S(x))+S(S(S(x)))=2016$ is
Since $x \equiv S(x) \equiv S(S(x)) \equiv S(S(S(x))) \pmod{9}$, and $2016 \equiv 0 \pmod{9}$, we have $4 S(x) \equiv 0 \pmod{9} \Rightarrow S(x) \equiv 0 \pmod{9}$. Also, $S(x) \leq 1+9+9+9+9=28$, so $S(x)=9,18,27$. When $S(x)=9$, $x+27=2016 \Rightarrow x=1989$, which is a contradiction; when $S(x)=18$, $x+36=2016 \Rightarrow x=1980$, which satisfies the condition; when $S(x)=27$, $x+45=2016 \Rightarrow x=1971$, which is a contradiction. In summary, the solution set of the equation $x+S(x)+S(S(x))+S(S(S(x)))=2016$ is $\{1980\}$.