Answer: A

Small Hint:

Cross-multiply and solve for $x$ in terms of $N$

Big Hint:

The equation reduces to $x=\frac{1}{2N+1};$ also solve this relation for $N$

Solution:

Cross-multiplication gives $$N(1+x)=(N+1)(1-x),$$ so $x(2N+1)=1$ and $$x=\frac1{2N+1}.$$ Every positive $N$ gives $0\lt x\lt1.$ Conversely, for any $0\lt x\lt1,$ $$N=\frac{1-x}{2x}>0,$$ so such an $N$ exists.

Thus, $x$ may be any positive value less than $1,$ and the correct answer is A.