This is just my first idea. Maybe you can refine it in something better, exploiting the fact that the values in between $1$, $3$, $5$ i.e., $2$ and $4$,
are also allowed because even. In the following I used a technique which should work for arbitrary values.
Substitute your integer variable $X$, wherever it appears, with the sum
where $x_1,x_2,x_3,x_4\ge0$, and $x_1,x_2,x_3\le1$ and the additional constraints
$x_1+x_2+x_3\le 1$, so no more than one value $x_1,x_2,x_3$ can be $1$.
$x_4 \le 50(1-x_1-x_2-x_3)$, because, if none of the $x_1,x_2,x_3$ are 1, then the bound is $x_4\le 50$, otherwise it is like $x_4\le 0$, so $x_4$ it is forced to be $0$.
$x_1+x_2+x_3+x_4\ge 1$, at least one value is taken, so $X$ cannot be $0$.
So, if $x_1=1$, the value is 1, if $x_2=1$, the value is $3$, if $x_3=1$ the value is $5$. If $x_4\ge1$ the value of $X$ is an even integer less or equal to $100$.