Consider all pairs of positive integers $(a,b)$, with $a<b$, such that 

$\sqrt{a} +\sqrt{b} = \sqrt{2,160}$

Determine all possible values of $a$.
1. Start with the given equation:
   \[
   \sqrt{a} + \sqrt{b} = \sqrt{2160}
   \]
   We need to find pairs of positive integers \((a, b)\) such that \(a < b\).

2. Isolate \(\sqrt{b}\):
   \[
   \sqrt{b} = \sqrt{2160} - \sqrt{a}
   \]

3. Square both sides to eliminate the square roots:
   \[
   b = (\sqrt{2160} - \sqrt{a})^2
   \]
   Expanding the right-hand side:
   \[
   b = 2160 + a - 2\sqrt{2160a}
   \]

4. For \(b\) to be an integer, \(2\sqrt{2160a}\) must be an integer. Let \(k = \sqrt{15a}\), then:
   \[
   2\sqrt{2160a} = 2\sqrt{2160 \cdot 15k^2} = 2 \cdot 24 \cdot 15k = 720k
   \]
   Therefore:
   \[
   b = 2160 + a - 720k
   \]

5. Since \(k = \sqrt{15a}\) must be an integer, \(15a\) must be a perfect square. Let \(15a = (15k)^2\), then:
   \[
   a = 15k^2
   \]

6. Substitute \(a = 15k^2\) into the equation for \(b\):
   \[
   b = 2160 + 15k^2 - 720k
   \]

7. Simplify the expression for \(b\):
   \[
   b = 15(144 + k^2 - 48k)
   \]
   \[
   b = 15(12 - k)^2
   \]

8. Since \(a < b\), we need \(15k^2 < 15(12 - k)^2\). Simplify:
   \[
   k^2 < (12 - k)^2
   \]
   \[
   k^2 < 144 - 24k + k^2
   \]
   \[
   0 < 144 - 24k
   \]
   \[
   24k < 144
   \]
   \[
   k < 6
   \]

9. Therefore, \(k\) can be any integer from 1 to 5. The pairs \((a, b)\) are:
   \[
   (a, b) \in \left\{ (15k^2, 15(12 - k)^2) : k \in \{1, 2, 3, 4, 5\} \right\}
   \]

10. Verify by direct substitution:
    - For \(k = 1\):
      \[
      a = 15 \cdot 1^2 = 15, \quad b = 15 \cdot 11^2 = 1815
      \]
      \[
      \sqrt{15} + \sqrt{1815} = \sqrt{2160}
      \]
    - For \(k = 2\):
      \[
      a = 15 \cdot 2^2 = 60, \quad b = 15 \cdot 10^2 = 1500
      \]
      \[
      \sqrt{60} + \sqrt{1500} = \sqrt{2160}
      \]
    - For \(k = 3\):
      \[
      a = 15 \cdot 3^2 = 135, \quad b = 15 \cdot 9^2 = 1215
      \]
      \[
      \sqrt{135} + \sqrt{1215} = \sqrt{2160}
      \]
    - For \(k = 4\):
      \[
      a = 15 \cdot 4^2 = 240, \quad b = 15 \cdot 8^2 = 960
      \]
      \[
      \sqrt{240} + \sqrt{960} = \sqrt{2160}
      \]
    - For \(k = 5\):
      \[
      a = 15 \cdot 5^2 = 375, \quad b = 15 \cdot 7^2 = 735
      \]
      \[
      \sqrt{375} + \sqrt{735} = \sqrt{2160}
      \]

The final answer is \(\boxed{15, 60, 135, 240, 375}\).