$\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}+ \dots+a_{18}x^{18}$
$$
Put $x=-1,$ we get
$$
\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.