Let $p$ be a prime satisfying $p^2\mid 2^{p-1}-1$, and let $n$ be a positive integer. Define
\[ f(x) = \frac{(x-1)^{p^n}-(x^{p^n}-1)}{p(x-1)}. \]
Find the largest positive integer $N$ such that there exist polynomials $g(x)$, $h(x)$ with integer coefficients and an integer $r$ satisfying $f(x) = (x-r)^N g(x) + p \cdot h(x)$.

[i]Proposed by Victor Wang[/i]
1. **Given Information and Definitions:**
   - Let \( p \) be a prime such that \( p^2 \mid 2^{p-1} - 1 \).
   - Define the function:
     \[
     f(x) = \frac{(x-1)^{p^n} - (x^{p^n} - 1)}{p(x-1)}.
     \]
   - We need to find the largest positive integer \( N \) such that there exist polynomials \( g(x) \), \( h(x) \) with integer coefficients and an integer \( r \) satisfying:
     \[
     f(x) = (x-r)^N g(x) + p \cdot h(x).
     \]

2. **Using Kummer's and Lucas' Theorems:**
   - By Kummer's theorem and Lucas' theorem, we can express:
     \[
     (x-1)^{p^n} - (x^{p^n} - 1) = \sum_{k=1}^{p^n-1} (-1)^{k+1} x^k \binom{p^n}{k}.
     \]
   - Modulo \( p^2 \), this sum simplifies to:
     \[
     \sum_{k=1}^{p-1} (-1)^{k+1} x^{p^{n-1} k} \binom{p}{k} = (x^{p^{n-1}} - 1)^p - (x^{p^n} - 1).
     \]

3. **Substitution and Simplification:**
   - Let \( x^{p^{n-1}} = X \). Then:
     \[
     \frac{(x-1)^{p^n} - (x^{p^n} - 1)}{p} \equiv \frac{(X-1)^p - (X^p - 1)}{p} \pmod{p}.
     \]
   - Define \( g(X) = \frac{(X-1)^p - (X^p - 1)}{p} \).

4. **Analyzing \( g(X) \):**
   - The second derivative of \( g(X) \) is:
     \[
     g''(X) = (p-1)((X-1)^{p-2} - X^{p-2}).
     \]
   - Since \( g''(X) \) has no solutions modulo \( p \), every root of \( g(X) \) must have a multiplicity at most 2.

5. **Wieferich Prime Condition:**
   - Given \( p \) is a Wieferich prime, we have:
     \[
     g(2) \equiv 0 \quad \text{and} \quad g'(2) = (2-1)^{p-1} - 2^{p-1} \equiv 0.
     \]
   - Hence, \( (X-2)^2 \) is a factor of \( g(X) \).

6. **Factorization and Polynomial Form:**
   - Let \( g(X) = (X-a_1) \cdots (X-a_m) R(X) \) where \( R(X) \) has no roots modulo \( p \).
   - Then:
     \[
     f(x) = \frac{(x^{p^{n-1}} - a_1) \cdots (x^{p^{n-1}} - a_m) R(x^{p^{n-1}})}{x-1}.
     \]
   - Modulo \( p \), this simplifies to:
     \[
     f(x) \equiv \frac{(x-a_1)^{p^{n-1}} \cdots (x-a_m)^{p^{n-1}} R(x^{p^{n-1}})}{x-1} \pmod{p}.
     \]

7. **Conclusion:**
   - Since \( x^{p^{n-1}} \equiv x \pmod{p} \) for all \( x \), the polynomial \( R(x^{p^{n-1}}) \) will have no roots modulo \( p \).
   - Therefore, the largest \( N \) is \( 2p^{n-1} \), which is obtained at \( r = 2 \) (or more).

The final answer is \( \boxed{ 2p^{n-1} } \).