Suppose that $f : Y \to X$ is a birational morphism, and $E$ is some effective exceptional divisor on $Y$. The negativity lemma says that $E \cdot C \leq 0$ for any curve $C$ contracted by $f$. I'm wondering about a sort of converse: suppose $C$ is a curve such that $E \cdot C < 0$. Must $C$ be contracted by $f$? Would any additional conditions guarantee that it is?



Curves negative for an exceptional divisor
No. $E\cdot C<0$ implies that $C\subseteq E$, but it may not be contracted.
In fact, if $E$ is not contracted to a point, then many curves in $E$ will be like that.


Assume that $f$ is the blow up of a smooth subvariety of a smooth variety $\Sigma\subset X$ and $E=f^{-1}(\Sigma)$. Then $f|\_E:E\to\Sigma$ is a $\mathbb P^n$-bundle and 
$\mathscr O\_Y(E)|\_E\simeq \mathscr O\_{E/\Sigma}(-1)$. In other words, $-E|\_E\sim H$, where $H$ is an $f$-ample effective divisor. Now if $C\subset E$ is such that $f|\_C$ is non-constant but $C\cap H\neq\emptyset$, then $E\cdot C<0$ but $C$ is not contracted.


To see that this can actually happen, just take $X$ to be a smooth threefold, $\Sigma\subset X$ a smooth proper curve. Then $E\to\Sigma$ is a ruled surface and $H\subset E$ is a section. If $C$ is another (multi-)section that intersects $H$, then these satisfy the above criterion.


**By the way...**
The negativity lemma does not imply what you are saying. If there are more than one exceptional divisors, say $E\_1$ and $E\_2$, then for any contracted curve chosen so that $C\_1\subseteq E\_1$ and $C\_1\not\subseteq E\_2$, it follows easily that $E\_2\cdot C\_1\geq 0$.


In particular, if say $X$ is a surface with an $A\_2$ singularity, then the exceptional set consists of two curves $E\_1$ and $E\_2$, they both have self-intersection $-2$, but $E\_1\cdot E\_2=1$.


