Let the $2009$-gon be $A_0A_1A_2...A_{2008}$ and let $\alpha_0 ,\alpha_1 ,..., \alpha_{2008} $ be the angles at $A_0,A_1,...,A_{2008}$, respectively . W.L.O.G let $A_0$ be the marked vertex. Note that the condition now translates to prove that the sum of angles at even indices must be equal to the sum of odd indices. If we remove any vertex with odd index $A_{2k+1}$ where $k \in \{1,2,3 ,...,1003 \}$ then the sum of angles colored with color $1$ and color $2$ are the same and we can get that : $$ \sum_{i=0}^{k} \alpha_{2i}+ \sum_{i=k+1}^{1003} \alpha_{2i+1}= \sum_{i=0}^{k-1} \alpha_{2i+1}+ \sum_{i=k}^{1004} \alpha_{2i} \;\;\;\;\;\;\;\; (1)$$ If we remove any vertex with even index $A_{2k}$ where $k \in \{1,2,3 ,...,1003 \}$ and do the same thing we get: $$\sum_{i=0}^{k-1} \alpha_{2i}+\sum_{i=k}^{1003} \alpha_{2i+1}=\sum_{i=0}^{k-1} \alpha_{2i+1}+ \sum_{i=k}^{1004} \alpha_{2i} \;\;\;\;\;\;\;\; (2)$$ $(1)-(2) $ gives us that $\boxed{\alpha_{2k}=\alpha_{2k+1}}$ for $k \in \{1,2,3 ,...,1003 \}$ If we remove $A_1$ we get that $$\alpha_0+\sum_{i=1}^{1003} \alpha_{2i+1}=\alpha_{2008}+\sum_{i=1}^{1003} \alpha_{2i} \Longleftrightarrow \boxed{\alpha_0=\alpha_{2008}}$$ If we remove $A_{2008}$ we get that $$\alpha_0+\sum_{i=1}^{1003} \alpha_{2i}=\alpha_1+\sum_{i=1}^{1003} \alpha_{2i+1} \Longleftrightarrow \boxed{\alpha_0=\alpha_1}$$ And we proved that $$\sum_{i=1}^{1004} \alpha_{2i}=\sum_{i=0}^{1003} \alpha_{2i+1} \;\;\;\; \blacksquare$$