Let's introduce the designations for the resistors and show the current directions on the circuit diagram (see the figure).
The ideal ammeter $A_{1}$ « short-circuits » points $A$ and $B$, therefore the total resistance of the circuit (between points $C$ and $D$) is
$$
R_{C D}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}+\frac{R_{3} R_{4}}{R_{3}+R_{4}}.
$$Then, the current flowing through ammeter $A_1$ is
$$
I_{0}=\frac{U\left(R_{1}+R_{2}\right)\left(R_{3}+R_{4}\right)}{R_{1} R_{2} R_{3}+R_{2} R_{3} R_{4}+R_{3} R_{4} R_{1}+R_{4} R_{1} R_{2}}. \quad (1)
$$The value of $\sum \Pi_{R}=R_{1} R_{2} R_{3}+R_{2} R_{3} R_{4}+R_{3} R_{4} R_{1}+R_{4} R_{1} R_{2}$ doesn't depend on the order of resistors and always equal to $50~
\Omega^{3}$.
Now, let's determine the current $I_0$ using the reading of ammeter $A_2$. From the given circuit it follows that
$$
\left.\begin{array}
{l}I_{1} R_{1}=I_{2} R_{2},
\\
\left(I_{1}-i\right) R_{3}=\left(I_{2}+i\right) R_{4}
\end{array}\right\} \Rightarrow
\\
\Rightarrow I_{1}=i \frac{R_{2}\left(R_{3}+R_{4}\right)}{R_{2} R_{3}-R_{1} R_{4}},
\\
I_{2}=i \frac{R_{1}\left(R_{3}+R_{4}\right)}{R_{2} R_{3}-R_{1} R_{4}}.
$$Thus,
$$
I_{0}=I_{1}+I_{2}=i \frac{\left(R_{1}+R_{2}\right)\left(R_{3}+R_{4}\right)}{R_{2} R_{3}-R_{1} R_{4}}. \quad (2)
$$Using $(1)$ and $(2)$, we obtain that
$$
R_{2} R_{3}-R_{1} R_{4}=\frac{i \sum \Pi_{R}}{U}=10~Ом^{2}. \quad (3)
$$This condition is satisfied if
1) $R_{2}=3~\Omega$, $R_{3}=4~\Omega$, $R_{1}=1~\Omega$, $R_{4}=2~\Omega$. In this case the resistance of the circuit is $R_{C D}=\frac{25}{12}~\Omega$ and the ammeter $A_{1}$ reads $I_{01}=4.8~А$.
2) $R_{2}=4~\Omega$, $R_{3}=3~\Omega$, $R_{1}=1~\Omega$, $R_{4}=2~\Omega$. In this case the resistance of the circuit is $R_{C D}=2~\Omega$ and the ammeter $A_{1}$ reads $I_{02}=5~А$.
Other permutations of the resistances satisfying condition (3) lead to the same results.
Thus, depending on the order in which the resistors are connected into the circuit, the reading of ammeter $A_{1}$ is:
$$
I_{01}=4.8~А \text { or } I_{02}=5~А.
$$