1. Set up the flow in the train frame

   Work in the frame of the train. Then:

   • The train is at rest.
   • Far away air approaches the train with speed $v_t$.
   • The air in front of the train is forced to flow backward through the gap between the train and tunnel wall.

   Given: $S_0 = 4S_t$ so the area available for backward flow is $S_0 - S_t = 4S_t - S_t = 3S_t.$

2. Use continuity equation

   Let the speed of air in the narrow gap region (relative to train) be $v$.

   The volume flow rate of air approaching the train from far ahead is $S_0 \, v_t.$

   This same air flows backward through the annular gap area $S_0-S_t$, so $S_0 v_t = (S_0-S_t)v.$

   Substituting $S_0=4S_t$: $4S_t v_t = 3S_t v$ $v = \frac{4}{3}v_t.$

3. Apply Bernoulli between far-away air and the gap

   In the train frame, far from the train:

   • pressure $= p_0$
   • speed $= v_t$

   In the gap region:

   • pressure $= p$
   • speed $= \dfrac{4}{3}v_t$

   Since the flow is steady and laminar, apply Bernoulli: $p_0 + \frac12 \rho v_t^2 = p + \frac12 \rho \left(\frac43 v_t\right)^2.$

   Therefore, $p_0 - p = \frac12 \rho \left[\left(\frac43 v_t\right)^2 - v_t^2\right].$

4. Simplify

   $p_0 - p = \frac12 \rho v_t^2\left(\frac{16}{9}-1\right)$ $= \frac12 \rho v_t^2\left(\frac{7}{9}\right)$ $= \frac{7}{18}\rho v_t^2.$

5. Compare with the given form

   Given: $p_0-p = \frac{7}{2N}\rho v_t^2.$

   So, $\frac{7}{2N} = \frac{7}{18}$ $2N = 18$ $N=9.$

6. Final answer

   $\boxed{9}$

The derived answer matches the stored correct answer.