Suppose that
\[ \frac {2x}{3} \minus{} \frac {x}{6}
\]is an integer. Which of the following statements must be true about $ x$?

$ \textbf{(A)}\ \text{It is negative.} \qquad \textbf{(B)}\ \text{It is even, but not necessarily a multiple of }3\text{.}$
$ \textbf{(C)}\ \text{It is a multiple of }3\text{, but not necessarily even.}$
$ \textbf{(D)}\ \text{It is a multiple of }6\text{, but not necessarily a multiple of }12\text{.}$
$ \textbf{(E)}\ \text{It is a multiple of }12\text{.}$
1. Start with the given expression:
   \[
   \frac{2x}{3} - \frac{x}{6}
   \]

2. Find a common denominator to combine the fractions. The least common multiple of 3 and 6 is 6:
   \[
   \frac{2x}{3} = \frac{2x \cdot 2}{3 \cdot 2} = \frac{4x}{6}
   \]
   So the expression becomes:
   \[
   \frac{4x}{6} - \frac{x}{6}
   \]

3. Combine the fractions:
   \[
   \frac{4x - x}{6} = \frac{3x}{6}
   \]

4. Simplify the fraction:
   \[
   \frac{3x}{6} = \frac{x}{2}
   \]

5. For \(\frac{x}{2}\) to be an integer, \(x\) must be even. This is because dividing an even number by 2 results in an integer.

6. Therefore, \(x\) must be even. There is no requirement for \(x\) to be negative, a multiple of 3, 6, or 12. 

Conclusion:
The correct statement about \(x\) is that it is even, but not necessarily a multiple of 3, 6, or 12.

The final answer is \(\boxed{\textbf{(B)}}\)