Is it possible for Cramer's V to be negative? To my understanding, Cramer's V cannot be negative because of the way we define it. But here's what I got from SAS:

 A: According to this SAS documentation, $V=\phi$ for $2\times2$ tables. It's not clear what your variable(s) is/are, but I'm guessing this explains your results.
However, note that Wikipedia defines Cramér's V in a way that should not allow negative numbers:

$$V = \sqrt{\frac{\varphi^2}{\min(k - 1,r-1)}} = \sqrt{ \frac{\chi^2/n}{\min(k - 1,r-1)}}$$
  where:
  
  
*
  
*$\varphi^2$ is the phi coefficient.
  
*$\chi^2$ is derived from Pearson's chi-squared test
  
*$n$ is the grand total of observations and
  
*$k$ being  the number of columns.
  
*$r$ being  the number of rows.
  

A negative $V$ cannot result from the formula above because of the square root operation. However, SAS' version of this formula preserves the sign of $\varphi$ for $2\times2$ tables, which allows $V$ to be identical to $\varphi$, as in your results. Presumably the denominator in the $V$ formula is 1, so $V=\sqrt{\varphi^2}$, but with the sign of $\varphi$ preserved...I suppose SAS does that because in this restricted circumstance you can tell which direction the association works in based on the sign of the coefficient and the orders of the group levels.
A: Values of $\phi$, which is a $2 \times 2$ version of $V$, can be negative, depending on the way you calculate them.
According to wikipedia, one of the ways is 
$$\phi = \frac{n_{11}n_{00}-n_{10}n_{01}}{\sqrt{n_{1\bullet} \cdot n_{0\bullet} \cdot  n_{\bullet0} \cdot  n_{\bullet1}}}$$
where 


*

*$n_{1\bullet} = n_{11} + n_{10}$  

*$n_{0\bullet} = n_{01} + n_{00}$ 

*$n_{\bullet0} = n_{10} + n_{00}$

*$n_{\bullet1} = n_{11} + n_{01}$


This one can easily be negative, and -1 would show "perfect disagreement". 
I'm not sure I've seen a formula with general $V$ calculated in a similar way, but I'd assume there must be one. 
References 


*

*http://en.wikipedia.org/wiki/Phi_coefficient
