Real numbers $a,b,c$ with $a\neq b$ verify $$a^2(b+c)=b^2(c+a)=2023.$$ Find the numerical value of $E=c^2(a+b)$.
Given the equations:
\[ a^2(b+c) = b^2(c+a) = 2023 \]

1. **Subtract the two given equations:**
   \[ a^2(b+c) - b^2(c+a) = 0 \]
   This simplifies to:
   \[ a^2b + a^2c - b^2c - b^2a = 0 \]
   Factor out common terms:
   \[ a^2b + a^2c - b^2c - b^2a = a^2b - b^2a + a^2c - b^2c = a(a-b)(b+c) + c(a^2 - b^2) = 0 \]
   Since \(a \neq b\), we can factor further:
   \[ (a-b)(ab + ac + bc + b^2) = 0 \]
   Since \(a \neq b\), it must be that:
   \[ ab + ac + bc + b^2 = 0 \]
   Simplifying, we get:
   \[ ab + ac + bc = 0 \]

2. **Express \(c\) in terms of \(a\) and \(b\):**
   From \(ab + ac + bc = 0\), we can solve for \(c\):
   \[ c(a + b) = -ab \]
   \[ c = -\frac{ab}{a + b} \]

3. **Substitute \(c\) back into the original equation:**
   Substitute \(c = -\frac{ab}{a + b}\) into \(a^2(b+c) = 2023\):
   \[ a^2 \left( b - \frac{ab}{a + b} \right) = 2023 \]
   Simplify the expression inside the parentheses:
   \[ a^2 \left( \frac{b(a + b) - ab}{a + b} \right) = 2023 \]
   \[ a^2 \left( \frac{ab + b^2 - ab}{a + b} \right) = 2023 \]
   \[ a^2 \left( \frac{b^2}{a + b} \right) = 2023 \]
   \[ \frac{a^2 b^2}{a + b} = 2023 \]

4. **Find \(E = c^2(a + b)\):**
   Substitute \(c = -\frac{ab}{a + b}\) into \(E\):
   \[ E = \left( -\frac{ab}{a + b} \right)^2 (a + b) \]
   \[ E = \frac{a^2 b^2}{(a + b)^2} (a + b) \]
   \[ E = \frac{a^2 b^2}{a + b} \]
   From the previous step, we know:
   \[ \frac{a^2 b^2}{a + b} = 2023 \]
   Therefore:
   \[ E = 2023 \]

The final answer is \(\boxed{2023}\).