smc

We compute the entire sum as $$\begin{array}{rl}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}& =\frac{60+30+20+15+12+10}{120}\\ & =\frac{147}{120}\\ & =\frac{49}{40},\end{array}$$ so the two terms to be removed must sum to $\frac{49}{40}-1=\frac{9}{40}$. That is, $\frac{9}{40}=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$ for some $x,y\in \{2,4,6,8,10,12\}$. Since $9$ and $40$ are coprime, we deduce that $xy$ is a multiple of $40$, so $x$ or $y$ must be a multiple of $5$. It follows that one of them must be $10$, and we obtain $$\begin{array}{rl}\frac{9}{40}-\frac{1}{10}& =\frac{9-4}{40}\\ & =\frac{5}{40}\\ & =\frac{1}{8},\end{array}$$ so the two terms we must remove are $\boxed{\frac{1}{8}\text{ and }\frac{1}{10}}$.