$ m$ and $ n$ are positive integers. In a $ 8 \times 8$ chessboard, $ (m,n)$ denotes the number of grids a Horse can jump in a chessboard ($ m$ horizontal $ n$ vertical or $ n$ horizontal $ m$ vertical ). If a $ (m,n) \textbf{Horse}$ starts from one grid, passes every grid once and only once, then we call this kind of Horse jump route a $ \textbf{H Route}$. For example, the $ (1,2) \textbf{Horse}$ has its $ \textbf{H Route}$. Find the smallest positive integer $ t$, such that from any grid of the chessboard, the $ (t,t\plus{}1) \textbf{Horse}$ does not has any $ \textbf{H Route}$.
To solve this problem, we need to determine the smallest positive integer \( t \) such that a \((t, t+1)\) Horse cannot complete an \( \textbf{H Route} \) on an \( 8 \times 8 \) chessboard. 

1. **Understanding the Horse's Moves:**
   A \((t, t+1)\) Horse can move in two ways:
   - \( t \) steps horizontally and \( t+1 \) steps vertically.
   - \( t+1 \) steps horizontally and \( t \) steps vertically.

2. **Analyzing the \((2, 3)\) Horse:**
   Let's consider \( t = 2 \). A \((2, 3)\) Horse can move:
   - 2 steps horizontally and 3 steps vertically.
   - 3 steps horizontally and 2 steps vertically.

3. **Checking the Reachability:**
   We need to check if a \((2, 3)\) Horse can visit every square on the \( 8 \times 8 \) chessboard exactly once. 

4. **Initial Position and Moves:**
   - From any corner, such as \((1,1)\), the possible moves are:
     - \((1+2, 1+3) = (3, 4)\)
     - \((1+3, 1+2) = (4, 3)\)
   - From \((1,7)\), the possible moves are:
     - \((1+2, 7+3) = (3, 10)\) (out of bounds)
     - \((1+3, 7+2) = (4, 9)\) (out of bounds)
   - From \((7,1)\), the possible moves are:
     - \((7+2, 1+3) = (9, 4)\) (out of bounds)
     - \((7+3, 1+2) = (10, 3)\) (out of bounds)
   - From \((7,7)\), the possible moves are:
     - \((7+2, 7+3) = (9, 10)\) (out of bounds)
     - \((7+3, 7+2) = (10, 9)\) (out of bounds)

5. **Cycle Formation:**
   - The moves from \((1,1)\) form a cycle with \((3,4)\), \((4,3)\), \((5,4)\), and \((4,5)\).
   - Similar cycles can be formed by translating these points to other parts of the board.

6. **Endpoints and Path Cutting:**
   - Since we only have two endpoints on the path, we can cut at most two of the cycles, which is not enough to cover the entire board.

7. **Conclusion:**
   - Therefore, a \((2, 3)\) Horse cannot complete an \( \textbf{H Route} \) on an \( 8 \times 8 \) chessboard.

The final answer is \( \boxed{ t = 2 } \).