Does there exist a four-digit positive integer with different non-zero digits, which has the following property: if we add the same number written in the reverse order, then we get a number divisible by $101$?

1. Let the four-digit number be represented as \(\overline{abcd}\), where \(a, b, c, d\) are distinct non-zero digits.
2. The number \(\overline{abcd}\) can be expressed as \(1000a + 100b + 10c + d\).
3. The reverse of this number, \(\overline{dcba}\), can be expressed as \(1000d + 100c + 10b + a\).
4. We need to check if the sum of these two numbers is divisible by 101:
   \[
   (1000a + 100b + 10c + d) + (1000d + 100c + 10b + a)
   \]
5. Simplifying the sum:
   \[
   1000a + 100b + 10c + d + 1000d + 100c + 10b + a = 1001a + 110b + 110c + 1001d
   \]
6. We need this sum to be divisible by 101:
   \[
   1001a + 110b + 110c + 1001d \equiv 0 \pmod{101}
   \]
7. Notice that \(1001 \equiv 1001 \mod 101\):
   \[
   1001 \equiv 1001 - 10 \cdot 101 = 1001 - 1010 = -9 \equiv 92 \pmod{101}
   \]
8. Therefore, the equation becomes:
   \[
   92a + 110b + 110c + 92d \equiv 0 \pmod{101}
   \]
9. Simplifying further:
   \[
   92(a + d) + 110(b + c) \equiv 0 \pmod{101}
   \]
10. Since \(110 \equiv 110 - 101 = 9 \pmod{101}\):
    \[
    92(a + d) + 9(b + c) \equiv 0 \pmod{101}
    \]
11. We can divide the entire equation by 9 (since 9 and 101 are coprime):
    \[
    92(a + d) \equiv -9(b + c) \pmod{101}
    \]
12. Since \(92 \equiv -9 \pmod{101}\):
    \[
    -9(a + d) \equiv -9(b + c) \pmod{101}
    \]
13. Dividing both sides by -9:
    \[
    a + d \equiv b + c \pmod{101}
    \]
14. This means that the sum of the first and last digits must be congruent to the sum of the second and third digits modulo 101.

The final answer is \( \boxed{ a + d \equiv b + c \pmod{101} } \)