Let $t$ and $n$ be fixed integers each at least $2$. Find the largest positive integer $m$ for which there exists a polynomial $P$, of degree $n$ and with rational coefficients, such that the following property holds: exactly one of \[ \frac{P(k)}{t^k} \text{ and } \frac{P(k)}{t^{k+1}} \] is an integer for each $k = 0,1, ..., m$.

[i]Proposed by Michael Kural[/i]
1. **Understanding the Problem:**
   We need to find the largest positive integer \( m \) such that there exists a polynomial \( P(x) \) of degree \( n \) with rational coefficients, where exactly one of \( \frac{P(k)}{t^k} \) and \( \frac{P(k)}{t^{k+1}} \) is an integer for each \( k = 0, 1, \ldots, m \).

2. **Reformulating the Condition:**
   The condition that exactly one of \( \frac{P(k)}{t^k} \) and \( \frac{P(k)}{t^{k+1}} \) is an integer can be interpreted using \( v_t \), the \( t \)-adic valuation. Specifically, this condition is equivalent to:
   \[
   v_t(P(k)) = k
   \]
   for each \( k = 0, 1, \ldots, m \).

3. **Constructing the Polynomial:**
   To construct such a polynomial \( P(x) \) of degree \( n \), we can use Lagrange Interpolation. Choose \( a_i \) such that \( v_t(a_i) = i \) for \( i = 0, 1, \ldots, n \). Then, construct \( P(x) \) such that:
   \[
   P(i) = a_i \quad \text{for all } i \leq n
   \]
   This ensures that \( v_t(P(i)) = i \) for \( i = 0, 1, \ldots, n \).

4. **Inductive Proof:**
   We need to show that \( m \) cannot exceed \( n \). We use induction on \( n \).

   - **Base Case:**
     For \( n = 1 \), consider a polynomial \( P(x) = a_0 + a_1 x \). We need \( v_t(a_0) = 0 \) and \( v_t(a_1) = 1 \). This polynomial satisfies the condition for \( k = 0 \) and \( k = 1 \). Thus, \( m = 1 \).

   - **Inductive Step:**
     Assume that for a polynomial of degree \( n \), \( m \) cannot exceed \( n \). Now consider a polynomial \( Q(x) \) of degree \( n+1 \) for which \( m \) can be more than \( n+1 \). Define the first-order difference \( \Delta_1(x) = Q(x+1) - Q(x) \). Note that \( \Delta_1(x) \) is a polynomial of degree \( n \).

     For \( \Delta_1(x) \), we have:
     \[
     v_t(\Delta_1(k)) = k \quad \text{for all } 0 \leq k \leq n+1
     \]
     This implies that \( \Delta_1(x) \) satisfies the condition for \( m = n+1 \), which contradicts our inductive hypothesis that \( m \) cannot exceed \( n \) for a polynomial of degree \( n \).

5. **Conclusion:**
   Therefore, the largest \( m \) for which there exists a polynomial \( P(x) \) of degree \( n \) with the given property is \( n \).

\(\blacksquare\)

The final answer is \( \boxed{ n } \)