4. Find all odd natural numbers $n$ for which

$$
n=2019 \underbrace{\varphi(\varphi(\ldots \varphi}_{10 \text { times }}(n) \ldots)) .
$$
4. In the task, we will use the following multiple times: if for natural numbers $a$ and $b$ it holds that $a \mid b$, then $\varphi(a) \mid \varphi(b)$ (indeed, if $a$ and $b$ are written in terms of their prime factorizations, $a=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$ and $b=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}} q_{1}^{\gamma_{1}} q_{2}^{\gamma_{2}} \cdots q_{l}^{\gamma_{l}}$, where $\beta_{i} \geqslant \alpha_{i}$ for $i=1,2, \ldots, k$, then we have $\frac{\varphi(b)}{\varphi(a)}=\prod_{i=1}^{k} p_{i}^{\beta_{i}-\alpha_{i}} \prod_{j=1}^{l} q_{j}^{\gamma_{j}-1}\left(q_{j}-1\right) \in \mathbb{N}$, which is what we claimed).

Let's introduce the notation $\varphi^{k}(n)=\varphi\left(\varphi^{k-1}(n)\right)$ and $\varphi^{1}(n)=\varphi(n)$. Denote $n=3^{\alpha} \cdot 673^{\beta} \cdot k$ for $(k, 4038)=1$ and $\alpha, \beta \geqslant 1$. First, let's prove $k=1$. If this is not the case, then $2 \mid \varphi(k)$. Then we have $\varphi(n)=3^{\alpha-1} \cdot 2 \cdot 673^{\beta-1} \cdot 672 \cdot \varphi(k)=2^{6} \cdot 3^{\alpha} \cdot 673^{\beta-1} \cdot 7 \cdot \varphi(k)$, and from this it follows that $2^{7} \cdot 3^{\alpha} \cdot 7 \mid \varphi(n)$. From this, we get $2^{8} \cdot 3^{\alpha}=\varphi\left(2^{7} \cdot 3^{\alpha} \cdot 7\right) \mid \varphi^{2}(n)$, and further $2^{8} \cdot 3^{\alpha-1} \mid \varphi^{3}(n)$, which implies $2^{8} \mid \varphi^{3}(n)$. Applying the function $\varphi$ 7 more times to both sides, we get $2 \mid \varphi^{10}(n)$, which leads to a contradiction.

Therefore, $n$ is written as $n=3^{\alpha} \cdot 673^{\beta}$.

Let's prove $\alpha=\beta=1$. Assume the opposite: at least one of the numbers $\alpha$ and $\beta$ is greater than 1. We will consider the case $\alpha \geqslant 2$, noting that $\beta \geqslant 2$ is solved analogously. Similarly to the first part, we calculate $\varphi(n)=2^{6} \cdot 3^{\alpha} \cdot 7 \cdot 673^{\beta-1}$. From this, we have in sequence $2^{6} \cdot 3^{\alpha} \cdot 7 \mid \varphi(n)$, then $2^{7} \cdot 3^{\alpha} \mid \varphi^{2}(n)$, then $2^{7} \cdot 3^{\alpha-1} \mid \varphi^{3}(n)$, and $2^{7} \cdot 3^{\alpha-2} \mid \varphi^{4}(n)$. From this, we conclude $2^{7} \mid \varphi^{4}(n)$, and then $2 \mid \varphi^{10}(n)$, which again leads to a contradiction. Therefore, the only solution is $n=2019$.