IRL ABooklet

Let the number of digits of $x$ be $d$, the leading digit of $x$ be $1 \le a \le 9$ and the remaining $d-1$ digits be $1 \le b \le 10^d - 1$. There is no solution with $d=1$, so we can assume $d > 1$. Then the equations to solve are:
$$\begin{align*} x &= 10^{d-1}a + b \\ 3x &= 10b + a \end{align*}$$
As $3x$ has the same number of digits as $x$, the leading digit of $x$ must be $a = 1, 2$ or $3$. Eliminating $x$, we obtain
$$a(3 \cdot 10^{d-1} - 1) = 7b.$$
As $b$ has $d-1$ digits, we have $b < 10^{d-1}$, which gives $(3a - 7)10^{d-1} < a$. Because $d > 1$ we can now exclude $a = 3$, so we must have $a = 1$ or $a = 2$. As $1 \le a \le 3$ it also follows that $7$ divides $3 \cdot 10^{d-1} - 1$, so that
$$10^d \equiv 3 \cdot 10^{d-1} \equiv 1 \pmod{7}.$$
Checking powers of 10 (mod 7), we confirm that 10 is a primitive root and so $10^d \equiv 1 \pmod 7$ if and only if $d \equiv 0 \pmod 6$. So let us write $d = 6c$ for an integer $c > 0$, then solutions must take the form:
$$b = \frac{a}{7} (3 \cdot 10^{6c-1} - 1) \quad \text{with } c > 0 \text{ and } a \in \{1, 2\}.$$
Adding the initial digit, we have:
$$\begin{align*} x &= 10^{6c-1}a + \frac{a}{7} (3 \cdot 10^{6c-1} - 1) \\ &= \frac{a}{7} (7 \cdot 10^{6c-1} + 3 \cdot 10^{6c-1} - 1) = \frac{a}{7} (10^{6c} - 1) \\ &= \frac{10^{6c} - 1}{10^6 - 1} \cdot \frac{10^6 - 1}{7} \cdot a = \frac{10^{6c} - 1}{10^6 - 1} \cdot 142857 \cdot a. \end{align*}$$
For any $c \ge 1$, we note that $(10^{6c} - 1)/(10^c - 1) = \sum_{k=0}^{c-1} 10^{6k}$ consists of $c$ ones, each separated by five zeros. Because $2 \cdot 142857 = 285714$ the solutions to the problem are numbers of the following form:

142857 repeated $c$ times and 285714 repeated $c$ times, $c \ge 1$.