For which $n$ do the equations have a solution in integers:  \begin{eqnarray*}x_1 ^2 + x_2 ^2 + 50 &=& 16x_1 + 12x_2 \\ x_2 ^2 + x_3 ^2 + 50 &=& 16x_2 + 12x_3 \\ \cdots \quad \cdots \quad \cdots & \cdots & \cdots \quad \cdots \\ x_{n-1} ^2 + x_n ^2 + 50 &=& 16x_{n-1} + 12x_n \\ x_n ^2 + x_1 ^2 + 50 &=& 16x_n + 12x_1 \end{eqnarray*}
1. We start by rewriting the given equations in a more manageable form. The original equations are:
   \[
   x_i^2 + x_{i+1}^2 + 50 = 16x_i + 12x_{i+1} \quad \text{for} \quad i = 1, 2, \ldots, n
   \]
   with \( x_{n+1} = x_1 \).

2. We can rearrange each equation to isolate the quadratic terms on one side:
   \[
   x_i^2 - 16x_i + x_{i+1}^2 - 12x_{i+1} + 50 = 0
   \]

3. Completing the square for \( x_i \) and \( x_{i+1} \):
   \[
   (x_i - 8)^2 - 64 + (x_{i+1} - 6)^2 - 36 + 50 = 0
   \]
   Simplifying, we get:
   \[
   (x_i - 8)^2 + (x_{i+1} - 6)^2 = 50
   \]

4. The equation \((x_i - 8)^2 + (x_{i+1} - 6)^2 = 50\) represents a circle centered at \((8, 6)\) with radius \(\sqrt{50}\). The integer solutions to this equation are the pairs \((x_i, x_{i+1})\) such that:
   \[
   (x_i - 8)^2 + (x_{i+1} - 6)^2 = 50
   \]

5. The possible pairs \((x_i - 8, x_{i+1} - 6)\) that satisfy this equation are:
   \[
   (7, 1), (-7, -1), (1, 7), (-1, -7), (5, 5), (-5, -5), (5, -5), (-5, 5)
   \]

6. Translating these back to the original variables, the possible pairs \((x_i, x_{i+1})\) are:
   \[
   (15, 7), (1, 5), (15, 5), (1, 7), (13, 11), (3, 1), (13, 1), (3, 11), (9, 13), (7, -1), (9, -1), (7, 13)
   \]

7. For the sequence \(x_1, x_2, \ldots, x_n\) to be valid, each pair \((x_i, x_{i+1})\) must be one of the above pairs, and \(x_{n+1} = x_1\). This implies that the sequence must alternate between two values to satisfy the conditions.

8. Therefore, each \(x_i\) must be either 1 or 7, and every pair of consecutive \(x_i, x_{i+1}\) must be different. This means the sequence must alternate between 1 and 7, forming a repeating pattern like \(1, 7, 1, 7, \ldots\) or \(7, 1, 7, 1, \ldots\).

9. For the sequence to close back on itself (i.e., \(x_{n+1} = x_1\)), \(n\) must be even. If \(n\) were odd, the sequence would not be able to alternate properly and return to the starting value.

Conclusion:
\[
\boxed{n \text{ must be even}}
\]