It is sufficient to prove that the following proposition $ P( n)$ is true for any $ n\in \mathbb{N}$ Proposition $ P( n)$ There does not exist $ x\in \mathbb{N}$ such that $ f( x) =f\left(\left( 10^{n} -1\right) /3\right) +1$. Proof For each $ n\in \mathbb{N}$, let $ g( n) =f\left(\left( 10^{n} -1\right) /3\right) +1$. It is not difficult to verify that $ P( 1)$ is true. Suppose that $ P( k)$ is true for some $ k\in \mathbb{N}$. Assume, for the sake of contradiction, $ P( k+1)$ is false. Then we can take $ x\in \mathbb{N}$ such that $ f( x) =g( k+1) \cdots ( \clubsuit )$. Note that $ g( k+1) =\left( 10^{k+1} -1\right) /3+3k+5\ \cdots ( \diamondsuit )$. If $ x\geq 10^{k+1}$, then $ 10^{k+1} < f( x)$, which is impossible. If $ x< 3\cdot 10^{k}$, then $ f( x) \leq \left( 3\cdot 10^{k} -1\right) +( 9k+2)$, which is impossible. So we must have $ 3\cdot 10^{k} \leq x< 10^{k+1}$. So we can take $ r\in \mathbb{N}$ such that $ x=3\cdot 10^{k} +r$ and $ r< 7\cdot 10^{k}$. Thus $ f( x) =x+S\left( 3\cdot 10^{k} +r\right) =x+S\left( 3\cdot 10^{k}\right) +S( r)$ $ =3\cdot 10^{k} +r+3+S( r) =f( r) +3\cdot 10^{k} +3$. It follows that $ f( x) =f( r) +3\cdot 10^{k} +3\ \cdots ( \heartsuit )$. From $ ( \diamondsuit )$, we have $ g( k+1) -g( k) =3\cdot 10^{k} +3\ \cdots ( \spadesuit )$. From $ ( \heartsuit )$,$ ( \spadesuit )$, and $ ( \clubsuit )$, we have $ 0=f( x) -g( k+1) =f( r) -g( k)$. This contradicts that $ P( k)$ is true. Therefore, $ P( k+1)$ is true. Hence $ P( n)$ is true for any $ n\in \mathbb{N}$ by mathematical induction. $ \blacksquare $

Let $ k$ be a positive integer and $ m=\left( 10^{k} +9k-1\right) /9$. Then we can verify that $ f\left( 10^{m} -2\right) =f\left( 10^{m} +10^{k} -3\right) =10^{m} +9m-3.$ Hence $ a=10^{m} +9m-3$ satisfies the condition where $ m=\left( 10^{k} +9k-1\right) /9$. $ \blacksquare $