A four-digit number's first two digits' sum is equal to the sum of the last two digits. The sum of the first and the last digit gives the third digit. The sum of the second and fourth digit is twice the sum of the first and the third. Which is this four-digit number?
A four-digit number in the decimal system, using the conventional notation, is:

$$
10^{3} a+10^{2} b+10 c+d
$$

where $a$ is between 1 and 9, and $b, c$, and $d$ are integers between 0 and 9.

According to the problem's conditions:

$$
a+b=c+d, \quad (1) \quad a+d=c, \quad (2) \quad b+d=2(a+c). \quad (3)
$$

This system of equations needs to be solved. From (1) and (2), we get $b=2d$, which means $d<5$. Substituting $c$ from (2) into (3), we get $b+d=4a+2d$. Substituting the previously found relationship for $b$, we get $d=4a$. Thus, the value of $a$ can be 1 or 2.

For $a=2, d=8$, then $b=2d=16$, which is not possible.

If $a=1, d=4$, then $b=8$, and from (2) $c=5$.

The found four-digit number is: 1854.

It can be seen that this number indeed satisfies the condition.