Suppose $X$ be a smooth projective curve over $\mathbb{C}$. Let $D$ be a divisor with $\deg D>0$ on $X$. Is it possible that $l(D)=0$, i.e. D is not linearly equivalent to an effective divisor? 



Does there exist a non effective divisor with positive degree?
If your curve is not rational or elliptic, i.e., $g\geq 2$ as Piotr commented, then there is a simple way to give an example:


Let $D=P\_1+P\_2-Q\_1$ for some $P\_1,P\_2, Q\_1\in X$ pairwise different points. Clearly $\deg D=1$, so if it were linearly equivalent to an effective divisor, it would have to be a point, say $Q\_2\in X$. It follows that then there exists a rational function on $X$ with zeroes at $P\_1$ and $P\_2$ and poles at $Q\_1$ and $Q\_2$. In other words, there exists a degree $2$ morphism $X\to \mathbb P^1$ with $P\_1$ and $P\_2$ mapping to the same point. Hence this works already if $X$ is not hyperelliptic. 


If $X$ is hyperelliptic, then this morphism is given by the basepoint-free linear system $|K\_X|$. In that case choose a new $P\_2$ (say $P\_2'$) that's different from the original $P\_2$. If $P\_1+P\_2'\in |K\_X|$, then $P\_2\sim P\_2'$, which would imply (similarly as above) that $X\simeq \mathbb P^1$ and if $P\_1+P\_2'\not\in |K\_X|$, then the above construction gives an example as desired.


**Remark** actually it is not important that the basepoint-free linear system in the second paragraph is $|K\_X|$, just that it is some linear system.


