Let $ S = a_1 + a_3 + \dots + a_{97} $ and $ T = a_2 + a_4 + \dots + a_{98} $. Then the given equation states that $ S + T = 137 $, and we want to find $ T $. We can build another equation relating $ S $ and $ T $: note that \[\begin{aligned} T-S &= (a_2-a_1) + (a_4-a_3) + \dots + (a_{98}-a_{97}) \\ &= \underbrace{1 + 1 + \dots + 1}_{49 \text{ times }} \\ &= 49 \end{aligned}\]since $ (a_n) $ has common difference $ 1 $. Then, adding the two equations $ S+T=137 $ and $ T-S=49 $, we get $ 2T=137+49=186 $, so $ T = \tfrac{186}{2} = \boxed{93} $.