Find the minimum and maximum values of the function
$$f(x,y,z,t)=\frac{ax^2+by^2}{ax+by}+\frac{az^2+bt^2}{az+bt},~(a>0,b>0),$$given that $x+z=y+t=1$, and $x,y,z,t\ge0$.
1. **Prove the lower bound \( f(x,y,z,t) \ge 1 \):**

   By the Cauchy-Schwarz inequality, we have:
   \[
   (ax^2 + by^2)(a + b) \ge (ax + by)^2
   \]
   Dividing both sides by \((a + b)(ax + by)\), we get:
   \[
   \frac{ax^2 + by^2}{ax + by} \ge \frac{ax + by}{a + b}
   \]
   Similarly, for the second term:
   \[
   \frac{az^2 + bt^2}{az + bt} \ge \frac{az + bt}{a + b}
   \]
   Adding these two inequalities, we obtain:
   \[
   f(x,y,z,t) = \frac{ax^2 + by^2}{ax + by} + \frac{az^2 + bt^2}{az + bt} \ge \frac{ax + by}{a + b} + \frac{az + bt}{a + b}
   \]
   Since \(x + z = 1\) and \(y + t = 1\), we have:
   \[
   f(x,y,z,t) \ge \frac{a(x + z) + b(y + t)}{a + b} = \frac{a \cdot 1 + b \cdot 1}{a + b} = 1
   \]

2. **Prove the upper bound \( f(x,y,z,t) \le 2 \):**

   Notice that:
   \[
   \frac{ax^2 + by^2}{ax + by} \le \max\{x, y\}
   \]
   and
   \[
   \frac{az^2 + bt^2}{az + bt} \le \max\{z, t\}
   \]
   Therefore:
   \[
   f(x,y,z,t) \le \max\{x, y\} + \max\{z, t\}
   \]
   Given \(x + z = 1\) and \(y + t = 1\), we can assume without loss of generality that \(x \ge y\). Then:
   \[
   \max\{x, y\} = x \quad \text{and} \quad \max\{z, t\} = \max\{1 - x, 1 - y\}
   \]
   Since \(x \ge y\), we have \(1 - y \ge 1 - x\), so:
   \[
   \max\{z, t\} = 1 - y
   \]
   Thus:
   \[
   f(x,y,z,t) \le x + (1 - y) = x + 1 - y
   \]
   Since \(x - y \le 1\), it follows that:
   \[
   f(x,y,z,t) \le 2
   \]

3. **Equality cases:**

   - For the lower bound \(f(x,y,z,t) = 1\), take \(x = y = z = t = \frac{1}{2}\):
     \[
     f\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \frac{a \left(\frac{1}{2}\right)^2 + b \left(\frac{1}{2}\right)^2}{a \cdot \frac{1}{2} + b \cdot \frac{1}{2}} + \frac{a \left(\frac{1}{2}\right)^2 + b \left(\frac{1}{2}\right)^2}{a \cdot \frac{1}{2} + b \cdot \frac{1}{2}} = 1
     \]

   - For the upper bound \(f(x,y,z,t) = 2\), take \(x = t = 1\) and \(y = z = 0\):
     \[
     f(1, 0, 0, 1) = \frac{a \cdot 1^2 + b \cdot 0^2}{a \cdot 1 + b \cdot 0} + \frac{a \cdot 0^2 + b \cdot 1^2}{a \cdot 0 + b \cdot 1} = 1 + 1 = 2
     \]

Thus, we have established that \(1 \le f(x,y,z,t) \le 2\) for all such \(x, y, z, t\).

\(\blacksquare\)

The final answer is \( \boxed{ 1 \le f(x,y,z,t) \le 2 } \)