For every integer $r > 1$ find the smallest integer $h(r) > 1$ having the following property: For any partition of the set $\{1, 2, . . ., h(r)\}$ into $r$ classes, there exist integers $a \geq 0, 1 \leq x \leq y$ such that the numbers $a + x, a + y, a + x + y$ are contained in the same class of the partition.
1. **Restate the problem and the goal:**
   We need to find the smallest integer \( h(r) > 1 \) such that for any partition of the set \(\{1, 2, \ldots, h(r)\}\) into \( r \) classes, there exist integers \( a \geq 0 \), \( 1 \leq x \leq y \) such that the numbers \( a + x \), \( a + y \), and \( a + x + y \) are contained in the same class of the partition.

2. **Reformulate the condition:**
   The condition can be rephrased as finding integers \( m \) and \( n \) in the same class such that \( m < n \leq 2m \). If such integers exist, we can set \( a = 2m - n \) and \( x = y = n - m \). Conversely, if no such integers exist, then elements of each class form an increasing sequence \( e_1 < e_2 < \ldots < e_k \) with \( e_{i+1} > 2e_i \) for every \( i \). This implies that in any triple of elements from the same class, the largest one is greater than the sum of the other two, and thus there is no triple \( a + x, a + y, a + x + y \) in the same class.

3. **Determine \( h(r) \):**
   We need to show that \( h(r) = 2r \).

4. **Prove \( h(r) = 2r \):**
   - **Upper Bound:**
     Consider the set \(\{1, 2, \ldots, 2r\}\). If this set is partitioned into \( r \) classes, by the pigeonhole principle, at least two elements from the set \(\{r, r+1, \ldots, 2r\}\) (which has size \( r+1 \)) must be in the same class. Let these elements be \( m \) and \( n \) with \( m < n \). Since \( n \leq 2r \) and \( m \geq r \), we have \( m < n \leq 2m \). Thus, the condition is satisfied.

   - **Lower Bound:**
     We need to show that \(\{1, 2, \ldots, 2r-1\}\) can be partitioned into \( r \) classes such that there are no \( m, n \) with \( m < n \leq 2m \) in any class. We can define such a partition recursively:
       - The elements \(\{r, r+1, \ldots, 2r-1\}\) go into \( r \) separate classes.
       - For every odd integer \( k \) in the interval \( 1 < k \leq 2r-1 \), the integer \(\frac{k-1}{2}\) goes into the same class as \( k \).

5. **Verify the partition:**
   - For \( r = 2 \), the set \(\{1, 2, 3\}\) can be partitioned into two classes: \(\{1, 3\}\) and \(\{2\}\). Here, there are no \( m, n \) with \( m < n \leq 2m \) in any class.
   - For \( r = 3 \), the set \(\{1, 2, 3, 4, 5\}\) can be partitioned into three classes: \(\{1, 3\}\), \(\{2, 5\}\), and \(\{4\}\). Again, there are no \( m, n \) with \( m < n \leq 2m \) in any class.

Thus, the smallest integer \( h(r) \) satisfying the given condition is \( 2r \).

The final answer is \( \boxed{ h(r) = 2r } \).