 Solution

Use the one-relation language $\{<\}$ and the total orderswhere the second is the order sum: every point in its natural-number block lies below every point in its integer block.

To show II wins every finite Ehrenfeucht-Fraïssé game, use the interval threshold strategy for discrete linear orders. With $k$ rounds left, corresponding open intervals and end rays between already matched points have either the same size below $h_k=2^k-1$, or both have at least $h_k$ elements. Their sizes may be infinite. The chosen-point correspondence is order preserving. Initially the unique interval is infinite in both orders, so the invariant holds.

If I chooses a repeated point, II repeats its counterpart. If I chooses in a short interval, II chooses the point with the same finite rank in the matching interval. In a long interval, let $h=h_{k-1}$. If one side of the chosen point has fewer than $h$ elements, reproduce that size in the other interval; the opposite sides then both have at least $h$ elements. Otherwise choose a response having at least $h$ elements on each side. Such a response exists because the interval has at least $2h+1=h_k$ elements. The endpoint and successor/predecessor properties of these two discrete orders allow the finite small-side counts to be reproduced. This establishes the invariant with $k-1$ rounds left and gives a finite-game winning strategy. Therefore II wins the timed game.

In the infinite Ehrenfeucht-Fraïssé game, I first chooses a point $b$ in the integer block of $N$. It has infinitely many predecessors. II must respond with some $m\in \mathbb{N}$, which has exactly $m$ predecessors. I then chooses $m+1$ distinct predecessors of $b$. To preserve order and equality, II would need $m+1$ distinct points below $m$, which is impossible. ThusI chooses the required finite length only after seeing II's first response, which explains why this does not contradict the timed-game result.