Sum of functions

The problem can be generalized to an existence of any $m\in { N }^{ * }$ which satisfies $f\left( m \right) =1$ . I will show that, in fact, $f\left( n \right) =1$ for any $n\in { N }^{ * }$ . Firstly, substituting n for m in the property we obtain $f\left( m+1 \right) =1$ , then putting $n=m+1$ we have $f\left( m+2 \right) =1$ , so by induction we deduce $f\left( n \right) =1$ for any $n\ge m$ . Now let's consider $A=\left\{ n|f(n)\neq 1,n\in { N }^{ * } \right\}$ . Suppose $A\neq \emptyset$ . We have $f\left( n+f\left( n \right) \right) =f\left( n \right)$ so for every $n\in A$ we have $n+f\left( n \right)\in A$ and by repeating this step we have $n+kf\left( n \right)\in A$ for any positive integer k. This means that there is $k\in { N }^{ * }$ such that $n+kf\left( n \right)\ge m$ so $n+kf\left( n \right)=1$ and this implies that the supposition is false. To conclude, we have $f\left( n \right) =1$ for any $n\in { N }^{ * }$ hence the answer is 2017 .

My solution is not a solid one, But a bit of guess worked,

It is if $f(n+f(n))=f(n)$ , also $f(2017)=1$ ,

$\implies$ $f(1+2017)=f(2017) = f(2018)$

$\implies$ $f(2018)=1$ .

Now say,

$f(2018+1)\implies f(2019) =f(2018)=f(2017)=1$

Similarly,

$\implies$ $f(2017)=f(2018)=f(2019)=f(2020)=\cdots=1$ . If a function has all values as $1$ after $2017$ , it is not possible that $f(2016) , f(2015) \cdots f(1)$ are other, but $1$ ,

Which proves $f(n)$ of above condition to be constant function

Constant Function The term means that whatever be the value of $n$ , $f(n)$ must be a constant $c$ , such that $f(n)=c$ for any $n \in \mathbb A$ ,such that $\mathbb A$ contains infinite or finite number of values.