For each positive integer $n$ let $s(n)$ denote the sum of the decimal digits of $n$. Find all pairs of positive integers $(a, b)$ with $a > b$ which simultaneously satisfy the following two conditions

$$a \mid b + s(a)$$
$$b \mid a + s(b)$$

[i]Proposed by Victor Domínguez[/i]
1. **Initial Observations:**
   - For any positive integer \( n \), the sum of its decimal digits \( s(n) \) satisfies \( s(n) \leq n \). This is because if \( n = \overline{n_k n_{k-1} \dots n_0} \) is the decimal representation of \( n \), then:
     \[
     n = 10^k n_k + 10^{k-1} n_{k-1} + \dots + 10 n_1 + n_0 \geq n_k + n_{k-1} + \dots + n_1 + n_0 = s(n)
     \]

2. **Analyzing the First Condition \( a \mid b + s(a) \):**
   - Given \( a > b \), we have \( b + s(a) \leq b + a < 2a \).
   - Since \( a \mid b + s(a) \) and \( 0 < b + s(a) < 2a \), the only possible value for \( b + s(a) \) that \( a \) can divide is \( a \). Therefore, we must have:
     \[
     a = b + s(a)
     \]

3. **Substituting into the Second Condition \( b \mid a + s(b) \):**
   - From \( a = b + s(a) \), substituting into the second condition gives:
     \[
     b \mid b + s(a) + s(b) \implies b \mid s(a) + s(b)
     \]

4. **Expressing \( b \) in Terms of \( a \):**
   - From \( a = b + s(a) \), we get:
     \[
     b = a - s(a)
     \]

5. **Analyzing the Expression \( n - s(n) \):**
   - We claim that if \( n \) has \( k+1 \) digits, then \( n - s(n) \geq 10^k - 1 \), with equality if and only if \( n = \overline{100 \dots 0 d} \) for some digit \( d \). This follows from:
     \[
     n - s(n) = (10^k - 1)n_k + (10^{k-1} - 1)n_{k-1} + \dots + (10^0 - 1)n_0 \geq 10^k - 1 + (n_{k-1} + \dots + n_1)
     \]

6. **Considering the Number of Digits in \( a \):**
   - Assume \( a \) has \( k+1 \) digits. From the claim, \( b \geq 10^k - 1 \). Since \( b < a \), both \( a \) and \( b \) have at most \( k+1 \) digits, and \( s(a), s(b) \leq 9(k+1) \).
   - Since \( b \mid s(a) + s(b) \), we get:
     \[
     10^k - 1 \leq 18(k+1)
     \]
   - This inequality is not true for \( k \geq 2 \). If \( k = 0 \), then \( a \) has exactly one digit, so \( a = s(a) \), and thus \( b = 0 \), which is not possible. Therefore, \( k = 1 \), and \( a \) has exactly two digits.

7. **Analyzing Two-Digit Numbers:**
   - Let \( a = \overline{cd} \), then \( a = 10c + d \) and \( b = a - s(a) = 10c + d - c - d = 9c \), so \( b \) is a multiple of 9.
   - Since \( s(a), s(b) \leq 18 \), we infer \( b \leq s(a) + s(b) \leq 36 \). This gives \( b < 99 \) and \( s(b) \neq 18 \). Since \( 9 \mid b \), we have \( s(b) = 9 \).

8. **Final Constraints:**
   - Substituting \( s(b) = 9 \) into \( b \leq s(a) + s(b) \):
     \[
     9c \leq c + d + 9 \implies 8c \leq d + 9
     \]
   - This is not possible if \( c \geq 3 \). Since \( 9 \mid 9c \) and \( 9c \mid c + d + 9 \), we have \( 9 \mid c + d \). The cases \( c = 2 \) and \( c = 1 \) give \( a = 27 \) and \( a = 18 \). These give the only solutions, which are the pairs \( (18, 9) \) and \( (27, 18) \).

The final answer is \( \boxed{ (18, 9) } \) and \((27, 18)\).