Let $M = \{(a,b,c,d)|a,b,c,d \in \{1,2,3,4\} \text{ and } abcd > 1\}$. For each $n\in \{1,2,\dots, 254\}$, the sequence $(a_1, b_1, c_1, d_1)$, $(a_2, b_2, c_2, d_2)$, $\dots$, $(a_{255}, b_{255},c_{255},d_{255})$ contains each element of $M$ exactly once and the equality \[|a_{n+1} - a_n|+|b_{n+1} - b_n|+|c_{n+1} - c_n|+|d_{n+1} - d_n| = 1\] holds.  If $c_1 = d_1 = 1$, find all possible values of the pair $(a_1,b_1)$.
1. **Understanding the Problem:**
   We are given a set \( M \) of quadruples \((a, b, c, d)\) where \(a, b, c, d \in \{1, 2, 3, 4\}\) and \(abcd > 1\). We need to find all possible values of the pair \((a_1, b_1)\) given that \(c_1 = d_1 = 1\) and the sequence \((a_n, b_n, c_n, d_n)\) satisfies the condition:
   \[
   |a_{n+1} - a_n| + |b_{n+1} - b_n| + |c_{n+1} - c_n| + |d_{n+1} - d_n| = 1
   \]
   for \(n \in \{1, 2, \dots, 254\}\).

2. **Counting the Elements in \(M\):**
   The total number of quadruples \((a, b, c, d)\) where \(a, b, c, d \in \{1, 2, 3, 4\}\) is \(4^4 = 256\). However, we exclude the quadruple \((1, 1, 1, 1)\) because \(abcd = 1\). Thus, the set \(M\) contains \(256 - 1 = 255\) elements.

3. **Parity Consideration:**
   Each step in the sequence changes exactly one of the coordinates by 1, which flips the parity of the sum \(a + b + c + d\). Since we start from \((a_1, b_1, c_1, d_1)\) and end at \((a_{255}, b_{255}, c_{255}, d_{255})\), the parity of the sum \(a + b + c + d\) must be the same for both the initial and final quadruples.

4. **Initial Quadruple Parity:**
   Given \(c_1 = d_1 = 1\), the sum \(a_1 + b_1 + 1 + 1 = a_1 + b_1 + 2\) must be odd. Therefore, \(a_1 + b_1\) must be odd.

5. **Possible Values for \((a_1, b_1)\):**
   The pairs \((a_1, b_1)\) where \(a_1 + b_1\) is odd are:
   \[
   \{(1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)\}
   \]

6. **Verification:**
   We need to verify that starting from any of these pairs, we can traverse all 255 elements of \(M\) while maintaining the given condition. Since each step changes the parity and we have an equal number of even and odd sums, starting from any of these pairs will allow us to cover all elements of \(M\).

The final answer is \( \boxed{ \{(1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)\} } \)