Find the number of ordered pairs $(a,b)$ such that $a,b \in \{10,11,\cdots,29,30\}$ and \\
$\hspace{1cm}$ $GCD(a,b)+LCM(a,b)=a+b$.
To solve the problem, we need to find the number of ordered pairs \((a, b)\) such that \(a, b \in \{10, 11, \ldots, 29, 30\}\) and \(\gcd(a, b) + \text{lcm}(a, b) = a + b\).

1. **Understanding the relationship between GCD and LCM:**
   We know that for any two integers \(a\) and \(b\),
   \[
   \gcd(a, b) \times \text{lcm}(a, b) = a \times b
   \]
   Given the condition \(\gcd(a, b) + \text{lcm}(a, b) = a + b\), we can use the above relationship to rewrite the equation:
   \[
   \gcd(a, b) + \frac{a \times b}{\gcd(a, b)} = a + b
   \]

2. **Simplifying the equation:**
   Let \(d = \gcd(a, b)\). Then we can write \(a = dk\) and \(b = dl\) where \(\gcd(k, l) = 1\). Substituting these into the equation, we get:
   \[
   d + \frac{d^2 k l}{d} = dk + dl
   \]
   Simplifying, we obtain:
   \[
   d + dkl = d(k + l)
   \]
   Dividing through by \(d\), we get:
   \[
   1 + kl = k + l
   \]
   Rearranging, we have:
   \[
   kl - k - l + 1 = 0
   \]
   Factoring, we get:
   \[
   (k-1)(l-1) = 1
   \]

3. **Solving the factored equation:**
   The equation \((k-1)(l-1) = 1\) has the integer solutions:
   \[
   (k-1, l-1) = (1, 1) \implies (k, l) = (2, 2)
   \]
   Therefore, \(k = l = 2\).

4. **Finding the pairs \((a, b)\):**
   Since \(a = dk\) and \(b = dl\) with \(k = l = 2\), we have:
   \[
   a = 2d \quad \text{and} \quad b = 2d
   \]
   Thus, \(a = b\) and both must be even numbers in the range \(\{10, 11, \ldots, 29, 30\}\).

5. **Counting the valid pairs:**
   The even numbers in the range \(\{10, 11, \ldots, 29, 30\}\) are:
   \[
   10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
   \]
   There are 11 even numbers. Each even number can pair with itself, giving us 11 pairs.

The final answer is \(\boxed{11}\)