7. Let $S(x)$ denote the sum of the digits of the natural number $x$. Then the solution set of the equation
$$
x+S(x)+S(S(x))=2013
$$

is $\qquad$ .
7. $\{1979,1985,1991,2003\}$.

Obviously, $x<2013$.
And $S(x)$ is at most $28$, $S(S(x))$ is at most 10, so $x$ is at least
$$
2013-38=1975 \text {. }
$$

Thus, $1975 \leqslant x<2013$.
Upon inspection, when $x=2003$,
$$
\begin{array}{l} 
S(2003)=5, S(S(2003))=5 \\
\Rightarrow 2003+5+5=2013 ;
\end{array}
$$

When $x=1991$,
$$
\begin{array}{l}
S(1991)=20, S(S(1991))=2 \\
\Rightarrow 1991+20+2=2013 ;
\end{array}
$$

When $x=1985$,
$$
\begin{array}{l}
S(1985)=23, S(S(1985))=5 \\
\Rightarrow 1985+23+5=2013 ;
\end{array}
$$

When $x=1979$,
$$
\begin{array}{l}
S(1979)=26, S(S(1979))=8 \\
\Rightarrow 1979+26+8=2013 ;
\end{array}
$$

Other natural numbers $x$ do not satisfy the equation.
Therefore, the solution set of the equation is
$$
\{1979,1985,1991,2003\} \text {. }
$$