Suppose $a,b,c,x,y,z$ are pairwisely different real numbers. How many terms in the following can be $1$ at most: 
$$\begin{aligned}
&ax+by+cz,&&&&ax+bz+cy,&&&&ay+bx+cz,\\
&ay+bz+cx,&&&&az+bx+cy,&&&&az+by+cx?
\end{aligned}$$
1. **Claim and Example:**
   We claim that the maximum number of terms that can be equal to 1 is 2. To illustrate this, consider the example where \(a = 1\), \(b = 2\), \(c = 3\), \(x = 4\), \(y = 27\), and \(z = -19\). In this case:
   \[
   ax + by + cz = 1 \quad \text{and} \quad ay + bz + cx = 1
   \]
   Therefore, it is possible to have two of the terms equal to 1.

2. **Proof by Contradiction:**
   We need to prove that it is impossible to have more than two terms equal to 1. Denote the expressions as follows:
   \[
   P(1) = ax + by + cz, \quad P(2) = ax + bz + cy, \quad P(3) = ay + bx + cz,
   \]
   \[
   P(4) = ay + bz + cx, \quad P(5) = az + bx + cy, \quad P(6) = az + by + cx
   \]

3. **Pairwise Comparison:**
   If \(P(1) = P(2) = 1\), then:
   \[
   ax + by + cz = 1 \quad \text{and} \quad ax + bz + cy = 1
   \]
   Subtracting these equations, we get:
   \[
   by + cz = bz + cy \implies b(y - z) = c(y - z)
   \]
   Since \(y \neq z\) and \(b \neq c\), this is a contradiction. Therefore, \(P(1)\) and \(P(2)\) cannot both be 1 simultaneously.

4. **Graph Construction:**
   Construct a graph with 6 vertices \(v_1, v_2, v_3, v_4, v_5, v_6\) corresponding to \(P(1), P(2), P(3), P(4), P(5), P(6)\). Connect \(v_i\) and \(v_j\) if and only if \(P(i)\) and \(P(j)\) can both be 1. From the pairwise comparison, we see that no two expressions sharing a common term can both be 1. This results in a graph with two disjoint triangles.

5. **Maximum Number of Terms Equal to 1:**
   From the graph, it is clear that at most 3 terms can be 1. However, we need to check if having 3 terms equal to 1 is possible.

6. **Case Analysis:**
   - **Case 1:** \(P(1) = ax + by + cz = 1\), \(P(4) = ay + bz + cx = 1\), \(P(5) = az + bx + cy = 1\)
     \[
     \begin{aligned}
     ax + by + cz &= 1, \\
     ay + bz + cx &= 1, \\
     az + bx + cy &= 1
     \end{aligned}
     \]
     Solving these equations simultaneously, we eliminate \(z\) and then \(y\) to find \(x\). This results in \(x = y = z = \frac{1}{a+b+c}\), which contradicts the given condition that \(x, y, z\) are pairwise different.

   - **Case 2:** \(P(2) = ax + bz + cy = 1\), \(P(3) = ay + bx + cz = 1\), \(P(6) = az + by + cx = 1\)
     This case is identical to the first case by swapping \(y\) and \(z\). Therefore, it also leads to a contradiction.

7. **Conclusion:**
   Since both cases lead to contradictions, it is impossible to have 3 of the terms equal to 1. Hence, the maximum number of terms that can be 1 is 2.

The final answer is \(\boxed{2}\)