A boy says to his friend: You can easily remember my phone number in the following way. There is a sequence of four natural numbers $a_{1} a_{2} a_{3} a_{4}$ in increasing order, where both $a_{2}$ and $a_{3}$ are the harmonic means of their neighbors, and when written next to each other, they form a six-digit number. This six-digit number is my phone number. Which one is it? ( $a_{2}$ is the harmonic mean of $a_{1}$ and $a_{3}$ if

$$
\frac{1}{a_{2}}=\frac{1}{2}\left(\frac{1}{a_{1}}+\frac{1}{a_{3}}\right)
$$
The number quadruple $a_{1} a_{2} a_{3} a_{4}$, when written consecutively, can only form a six-digit number if $a_{1}$ and $a_{2}$ are both single-digit. Express $a_{3}$ and $a_{4}$ using these. From the rearrangement of equation (1), we get

$$
a_{3}=\frac{a_{1} a_{2}}{2 a_{1}-a_{2}}
$$

and using this,

$$
a_{4}=\frac{a_{1} a_{2}}{3 a_{1}-2 a_{2}}
$$

According to the problem's requirement, $00
$$

which means

$$
3 a_{1}-2 a_{2} \geq 1
$$

and the condition $a_{3}<a_{4}$ implies

$$
2 a_{1}-a_{2} \geq 2
$$

From inequality (3), it follows that $a_{2}<\frac{3}{2} a_{1}$. Therefore,

$$
a_{1}<a_{2}<\frac{3}{2} a_{1}
$$

Since $a_{1}$ and $a_{2}$ are integers, inequality (6) can only hold if $a_{1} \geq 3$ and $a_{2} \geq 4$. Let the greatest common divisor of $a_{1}$ and $a_{2}$ be $d$, and $a_{1}=a_{1}^{\prime} d, a_{2}=a_{2}^{\prime} d$. Then

$$
a_{3}=\frac{a_{1}^{\prime} a_{2}^{\prime} d}{2 a_{1}^{\prime}-a_{2}^{\prime}} \quad \text { and } \quad a_{4}=\frac{a_{1}^{\prime} a_{2}^{\prime} d}{3 a_{1}^{\prime}-2 a_{2}^{\prime}}
$$

Similarly to inequality (6), we get

$$
a_{1}^{\prime}<a_{2}^{\prime}<\frac{3}{2} a_{1}^{\prime}
$$

and

$$
a_{1}^{\prime} \geq 3 \text { and } a_{2}^{\prime} \geq 4
$$

From this, it follows that $d \leq 2$, because if $d \geq 3$, then $a_{2}$ would already be a two-digit number.

Now we show that $a_{1}$ and $a_{2}$ cannot be coprime, i.e., $d \geq 2$; from which it follows that $d=2$. Suppose, indeed, that $a_{1}$ and $a_{2}$ are coprime. Then $a_{1}$ and $2 a_{1}-a_{2}$ are also coprime. Therefore, $a_{3}$ can only be an integer if $\frac{a_{2}}{2 a_{1}-a_{2}}$ is an integer. If $\frac{a_{2}}{2 a_{1}-a_{2}}$ is an integer, then $\frac{a_{2}}{2 a_{1}-a_{2}}+1=\frac{2 a_{1}}{2 a_{1}-a_{2}}$ is also an integer. But this can only be if $\frac{2}{2 a_{1}-a_{2}}$ is also an integer, because $a_{1}$ and $2 a_{1}-a_{2}$ are coprime by our assumption. This last quotient, considering inequality (5), can only be an integer if $2 a_{1}-a_{2}=2$. In this case, $3 a_{1}-2 a_{2}=1$, from which $a_{1}=3, a_{2}=4$, $a_{3}=6$ and $a_{4}=12$. This number quadruple forms only a five-digit number. Therefore, the number quadruple that satisfies the problem's requirements can only be found when $d=2$. In this case, due to inequality (8),

$$
6 \leq a_{1}<a_{2}<10
$$

and because $d=2$, $a_{1}$ and $a_{2}$ are both even. Therefore, $a_{1}$ can only be 6, and in this case $a_{2}=8$. From this, we get $a_{3}=12$, and $a_{4}=24$. The number quadruple $6,8,12,24$ is thus the only one that satisfies the problem's requirements.