1) Recall that for any $a$, the integers $a$ and $S(a)$ are congruent modulo 9. Indeed, if $\overline{a_{k} \cdots a_{1} a_{0}}$ is the decimal representation of $a$, then since $10 \equiv 1[9]$, we have for all $j \geqslant 0: 10^{j} \equiv 1^{j}=1[9]$, thus $a=\sum_{j=0}^{k} a_{j} 10^{j} \equiv \sum_{j=0}^{k} a_{j}=S(a)$ [9].

Suppose, for the sake of contradiction, that there exist $a$ and $b$ as described in the statement. Modulo 9, we have

$$\begin{aligned} a & \equiv S(a) \equiv 2013 \equiv 6 \\ b & \equiv S(b) \equiv 2013 \equiv 6 \\ a+b & \equiv S(a+b) \equiv 2013 \equiv 6 \end{aligned}$$

Adding the first two congruences and subtracting the third, we get $0 \equiv 6+6-6=6 [9]$. Impossible.
2) We note that $2016=9 \times 224$, so we can take $\mathrm{a}=\mathrm{b}=9090 \cdots 09$ where the digit 9 appears 224 times, and $a+b=1818 \cdots 18$ where the pattern 18 appears 224 times.
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