Four positive integers $a, b, c, d$ satisfy the condition: $a < b < c < d$. For what smallest possible value of $d$ could the following condition be true: the arithmetic mean of numbers $a, b, c$ is twice smaller than the arithmetic mean of numbers $a, b, c, d$?
1. Given the condition that the arithmetic mean of numbers \(a, b, c\) is twice smaller than the arithmetic mean of numbers \(a, b, c, d\), we can write this as:
   \[
   \frac{a + b + c}{3} = \frac{1}{2} \left( \frac{a + b + c + d}{4} \right)
   \]

2. Simplify the equation:
   \[
   \frac{a + b + c}{3} = \frac{1}{2} \cdot \frac{a + b + c + d}{4}
   \]
   \[
   \frac{a + b + c}{3} = \frac{a + b + c + d}{8}
   \]

3. Multiply both sides by 24 to clear the denominators:
   \[
   8(a + b + c) = 3(a + b + c + d)
   \]

4. Distribute and simplify:
   \[
   8a + 8b + 8c = 3a + 3b + 3c + 3d
   \]
   \[
   5a + 5b + 5c = 3d
   \]

5. Factor out the common term on the left side:
   \[
   5(a + b + c) = 3d
   \]

6. Solve for \(d\):
   \[
   d = \frac{5}{3}(a + b + c)
   \]

7. Since \(a, b, c, d\) are positive integers and \(a < b < c < d\), we need to find the smallest possible value of \(d\). The smallest values for \(a, b, c\) that satisfy \(a < b < c\) are \(1, 2, 3\).

8. Substitute \(a = 1\), \(b = 2\), and \(c = 3\) into the equation:
   \[
   d = \frac{5}{3}(1 + 2 + 3) = \frac{5}{3} \cdot 6 = 10
   \]

9. Verify that \(d = 10\) satisfies the condition \(a < b < c < d\):
   \[
   1 < 2 < 3 < 10
   \]

10. Check the arithmetic mean condition:
    \[
    \frac{1 + 2 + 3}{3} = 2
    \]
    \[
    \frac{1 + 2 + 3 + 10}{4} = 4
    \]
    \[
    2 = \frac{1}{2} \cdot 4
    \]

Thus, the smallest possible value of \(d\) is indeed \(10\).

The final answer is \(\boxed{10}\).