We know the following about the digits of a four-digit number:

I. The sum of the first and second (thousands and hundreds place) digits is equal to the sum of the last two digits.

II. The sum of the second and fourth digits is equal to twice the sum of the first and third digits.

III. Adding the first and fourth digits gives the third digit.

IV. Subtracting the first digit from the sum of the second and third digits gives three times the last digit.

Can this four-digit number be uniquely determined from these conditions?
Let the digits of the four-digit number be denoted by $a, b, c, d$ respectively, where $a$ is in the thousands place. Since it is a real four-digit number, $a \neq 0$ and $a \leq 9$, and $0 \leq b, c, d \leq 9$. Writing the conditions of the problem as equations, we get:

$$
a+b=c+d, \text { (I.) } \quad b+d=2(a+c), \text { (II.) } \quad a+d=c, \text { (III.) } \quad b+c-a=3 d \text { (IV.) }
$$

a system of equations.

Substituting $c$ from the III. equation into the IV., we get that $b=2 d$, and substituting this and $c$ from III. into II., we get $d=4 a$.

Substituting the values of $b, c$, and $d$ obtained into I., we arrive at the identity $a+2 d=a+2 d$, which means that our equations are not independent of each other.

Indeed, if we add the corresponding sides of equations I. and IV., after simplification, we get $2 c=2 a+2 d$, which is twice the III. equation.

Let's try to solve the system of equations based on our current knowledge.

We know that $d=4 a$, and $b=2 d=8 a$; it follows that $a$ can only be 1, since $b \leq 9$. If $a=1$, then $b=8$ and $d=4$, and since $a+b=c+d, c=5$ is the only possible value. The sought four-digit number is therefore uniquely determined, and it is 1854. It is easy to verify that it indeed satisfies the conditions.