Solve the following system of equations:

$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=50 \\
x_{1}^{2}-x_{2}^{2}+x_{3}^{2}-x_{4}^{2}=-24 \\
x_{1} x_{3}=x_{2} x_{4} \\
x_{1}-x_{2}+x_{3}+x_{4}=0
\end{gathered}
$$
From the sum of the first two equations, we get $x_{1}^{2}+x_{3}^{2}=13$, and from the difference of the first and second equations, we get $x_{2}^{2}+x_{4}^{2}=37$.

Rearranging the fourth equation, we obtain $x_{1}+x_{3}=x_{2}-x_{4}$. Squaring this, we get

$$
x_{1}^{2}+x_{3}^{2}+2 x_{1} x_{3}=x_{2}^{2}+x_{4}^{2}-2 x_{2} x_{4},
$$

from which, using the values found for the sums of squares,

$$
2 x_{1} x_{3}+2 x_{2} x_{4}=24
$$

According to the third equation, therefore, $x_{1} x_{3}=x_{2} x_{4}=6$. From this, we obtain the following system of equations:
I. $\mathrm{x}_{1}^{2}+\mathrm{x}_{3}^{2}=13$
II. $\mathrm{x}_{2}^{2}+\mathrm{x}_{4}^{2}=37$
$x_{1} x_{3}=6$
$x_{2} x_{4}=6$.

From I, $\left|x_{1}+x_{3}\right|=5$ and $\left|x_{1}-x_{3}\right|=1$, and from II, $\left|x_{2}+x_{4}\right|=7$ and $\left|x_{2}-x_{4}\right|=5$.

If we could independently choose the sign of the quantities within the absolute value signs, the system of equations would have $2^{4}=16$ solutions. However, since $x_{1}+x_{3}=x_{2}-x_{4}$, the signs of these two quantities must be the same, so the number of solutions is 8. We have summarized these in the table below:

| $x_{1}+x_{3}$ | 5 | 5 | 5 | 5 | -5 | -5 | -5 | -5 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{1}-x_{3}$ | 1 | -1 | 1 | -1 | 1 | -1 | 1 | -1 |
| $x_{2}+x_{4}$ | 7 | 7 | -7 | -7 | 7 | 7 | -7 | -7 |
| $x_{2}-x_{4}$ | 5 | 5 | 5 | 5 | -5 | -5 | -5 | -5 |
| $x_{1}$ | 3 | 2 | 3 | 2 | -2 | -3 | -2 | -3 |
| $x_{2}$ | 6 | 6 | -1 | -1 | 1 | 1 | -6 | -6 |
| $x_{3}$ | 2 | 3 | 2 | 3 | -3 | -2 | -3 | -2 |
| $x_{4}$ | 1 | 1 | -6 | -6 | 6 | 6 | -1 | -1 |

Since the obtained solutions consist of the roots of I and II, the first three equations of the original system are satisfied by them, and the fourth equation is satisfied due to the choice $x_{1}+x_{3}=x_{2}-x_{4}$. This completes the solution of the problem.