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 definnite. 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 tthe 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 $E_2(a_1E_1+a_2E_2)<0$.

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 $E_2(a_1E_1+a_2E_2)<0$.