Let $f$ be a potential solution.
Let $a \in \mathbb{N}^{*}$ such that $f(a)=1$.
Then $f(a+1)=f(a+f(a))=f(a)=1$.
Since $f(2013)=1$, we deduce by induction that $f(n)=1$ for all integers $\mathfrak{n} \geqslant 2013$.
On the other hand, suppose that $a \geqslant 2$ is an integer such that $f(n)=1$ for all $n \geqslant a$.
Then, $a-1 \geqslant 1$ and $f(a-1) \geqslant 1$, so $a-1+f(a-1) \geqslant a$ and thus from the first paragraph $f(a-1)=f(a-1+f(a-1))=1$.

This ensures that $f(n)=1$ for all $n \geqslant a-1$.
Since we have seen that $f(n)=1$ for all $n \geqslant 2013$, we deduce by downward induction that $f(n)=1$ for all $n \geqslant 1$.

Conversely, it is clear that the constant function $\mathrm{f}: \mathrm{n} \longmapsto 1$ is a solution to the problem.
Finally, the only solution is the constant function $\mathrm{f}: \mathfrak{n} \longmapsto 1$.
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