Consider $h+1$ chess boards. Number the squares of each board from 1 to 64 in such a way that when the perimeters of any two boards of the collection are brought into coincidence in any possible manner, no two squares in the same position have the same number. What is the maximum value of $h?$
1. **Understanding the Problem:**
   We need to number the squares of \( h+1 \) chess boards such that when the perimeters of any two boards are aligned, no two squares in the same position have the same number. Each chessboard has 64 squares, and we need to ensure that the numbers 1 to 64 are used in such a way that the condition is satisfied.

2. **Analyzing the Constraints:**
   If \( h \geq 16 \), then \( h+1 \geq 17 \). This means we have 17 boards. Each board has 8 corners, so the total number of corners is \( 17 \times 4 = 68 \). Since there are only 64 unique numbers available, it is impossible to assign numbers to the corners of 17 boards without repeating some numbers. This leads to a contradiction.

3. **Maximum Value of \( h \):**
   Therefore, the maximum value of \( h \) must be less than 16. We need to check if \( h = 15 \) is possible.

4. **Constructing the Solution for \( h = 15 \):**
   Let's denote the sets of squares in specific positions on each board as \( A, B, C, D, E, F, G \). The positions are as follows:
   ```
   ABCDEFGA
   G......B
   F......C
   E......D
   D......E
   C......F
   B......G
   AGFEDCBA
   ```

5. **Assigning Numbers to the Sets:**
   - For board \( i+1 \), assign the numbers \( 4i+1, 4i+2, 4i+3, 4i+4 \) to set \( A \).
   - For board \( i+2 \), assign the numbers \( 4i+1, 4i+2, 4i+3, 4i+4 \) to set \( B \).
   - Continue this pattern until board \( i+7 \) for set \( G \).

   This ensures that the perimeter of each board has no coincidental numbers in any position when aligned with another board.

6. **Reflection Consideration:**
   If the boards can be reflected, the maximum value of \( h \) is reduced to 7. This is because reflecting the boards introduces additional constraints, making it harder to avoid coincidental numbers.

The final answer is \( \boxed{ h = 15 } \) (without reflection) and \( h = 7 \) (with reflection).