Probably this is a very easy question. Let $f:X\rightarrow S$ be a resolution of a projective surface such that
$$K\_X = f^{\*}K\_S+\sum\_ia\_iE\_i$$
with $a\_i>0$. By Grauert-Mumford theorem the intersection matrix of the $E\_i$'s is negative definite. I found the following claim: there exists an $E\_j$ such that
$$E\_j\cdot(\sum\_ia\_iE\_i)<0.$$
Why is this true?



Intersection Matrix of a resolution
You can argue also in the following way (let us do the case of two componets $E\_1$, $E\_2$ for simplicity of notation. The general case will be clear):


the intersection matrix
$$
I = \left(\begin{array}{cc}
E\_1^2 & E\_1E\_2 \\ 
E\_1E\_2 & E\_2^2
\end{array}\right)
$$
is negative definite. In particular if you take the vector $a=(a\_1,a\_2)$ you get
$$a\cdot I\cdot a^{t} = a\_1^2E\_1^2+2a\_1a\_2E\_1E\_2+a\_2^2E\_2^2 <0.$$
On the other hand 
$$a\_1^2E\_1^2+2a\_1a\_2E\_1E\_2+a\_2^2E\_2^2 = a\_1E\_1(a\_1E\_1+a\_2E\_2)+a\_2E\_2(a\_1E\_1+a\_2E\_2)<0.$$
Since $a\_1,a\_2>0$ the last inequality yields either $E\_1(a\_1E\_1+a\_2E\_2)<0$ or $E\_2(a\_1E\_1+a\_2E\_2)<0$.


