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 $\qquad$
$-1 .\{1980\}$.
It is easy to see that $x<2016$.
Note that, the sum of the digits of natural numbers less than 2016 is at most 28, for example, $S(1999)=28$, which indicates,
$$
S(x) \leqslant 28 \text {. }
$$

Furthermore, $S(S(x)) \leqslant S(19)=10$.
Finally, $S(S(S(x))) \leqslant 9$.
From the equation we get
$$
\begin{array}{l}
x=2016-S(x)-S(S(x))-S(S(S(x))) \\
\geqslant 2016-28-10-9=1969 .
\end{array}
$$

Thus, $x \in\{1969,1970, \cdots, 2015\}$.
Also, $x, S(x), S(S(x)), S(S(S(x)))$ have the same remainder when divided by 9, and 2016 has a remainder of 0 when divided by 9, so each number has a remainder of 0 when divided by 9.

Therefore, $x$ can only be $1971, 1980, 1989, 1998, 2007$.
Upon inspection, only $1980+18+9+9=2016$. Therefore, the solution set of the equation is $\{1980\}$.