Let $\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}$ $+\ldots+a_{18} x^{18} \cdot$ Then

$\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}+ \dots+a_{18}x^{18}$ $$ Put $x=-1,$ we get $$ \begin{array}{l} (1-1+1)^{9}=a_{0}-a_{1}+a_{2}+\ldots+a_{18} \\ \Rightarrow 1=a_{0}-a_{1}+a_{2}+\ldots+a_{18} \end{array} $$ Put $x=1,$ we get $$ (1+1+1)^{9}=a_{0}+a_{1}+a_{2}+\ldots+a_{14} $$ $\Rightarrow 3^{9}=a_{0}+a_{1}+a_{2}+\ldots+a_{18}$ On adding Eqs. (i) and (ii), we get $$ \begin{array}{c} 3^{9}+1=2\left(a_{0}+a_{2}+\ldots+a_{10}\right) \\ a_{0}+a_{2}+a_{4}+\ldots+a_{18}=\frac{3^{0}+1}{2} \\ =\frac{19683+1}{2} \\ =\frac{19684}{2} \end{array} $$ $=9842$ which is even number.