We will prove that for every $k \in \mathbb{N}$ there exists $n_k$, such that $ka_{n_k}=n_k$. Essentially, we use induction. Let $b_n=n-a_n$. It is obvious that $b_n>0$ for $n$ large enough. Since $b_1 \leq 0$, and $b_n-b_{n-1} \leq 1$, for some $n_1$ we must have $b_{n_1}=0 \implies a_{n_1}=n_1$. For induction step, suppose $n_{k-1}$ satisfies $(k-1)a_{n_{k-1}}=n_{k-1}$. If we denote $c_{n}=n-ka_n$, then $c_{n}>0$ for large $n$. Also $c_{n_{k-1}}=n_{k-1}-ka_{n_{k-1}}=-a_{n_{k-1}}<0$ and $c_{n}-c_{n-1} \leq 1$, so for some $n_k>n_{k-1}$ we must have $c_{n_k}=0 \implies ka_{n_k}=n_k$.