jmc

Let the integer have digits $a$, $b$, and $c$, read left to right. Because $1\le a<b<c$, none of the digits can be zero and $c$ cannot be 2. If $c=4$, then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ choices for $a$ and $b$, and for each choice there is one acceptable order. Similarly, for $c=6$ and $c=8$ there are, respectively, $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ and $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ choices for $a$ and $b$. Thus there are altogether $3+10+21=\boxed{34}$ such integers.