Find all positive integer solutions $(a, b, c)$ to the function $a^{2} + b^{2} + c^{2} = 2005$, where $a \leq b \leq c$.
To find all positive integer solutions \((a, b, c)\) to the equation \(a^2 + b^2 + c^2 = 2005\) where \(a \leq b \leq c\), we start by considering the equation modulo 8. 

1. **Consider the equation modulo 8:**
   \[
   a^2 + b^2 + c^2 \equiv 2005 \pmod{8}
   \]
   Since \(2005 \equiv 5 \pmod{8}\), we need to find the possible values of \(a^2, b^2, c^2 \pmod{8}\). The possible values of a square modulo 8 are \(0, 1, 4\). Therefore, we need to find combinations of these values that sum to 5 modulo 8.

2. **Possible combinations:**
   \[
   \begin{aligned}
   &0 + 1 + 4 \equiv 5 \pmod{8}, \\
   &1 + 0 + 4 \equiv 5 \pmod{8}, \\
   &4 + 1 + 0 \equiv 5 \pmod{8}.
   \end{aligned}
   \]
   Without loss of generality, assume \(a^2 \equiv 0 \pmod{8}\), \(b^2 \equiv 1 \pmod{8}\), and \(c^2 \equiv 4 \pmod{8}\).

3. **Express \(a, b, c\) in terms of new variables:**
   \[
   \begin{aligned}
   &a = 4a_1, \\
   &b = 2b_1 + 1, \\
   &c = 4c_1 + 2.
   \end{aligned}
   \]
   Substituting these into the original equation:
   \[
   (4a_1)^2 + (2b_1 + 1)^2 + (4c_1 + 2)^2 = 2005
   \]
   Simplifying:
   \[
   16a_1^2 + 4b_1^2 + 4b_1 + 1 + 16c_1^2 + 16c_1 + 4 = 2005
   \]
   \[
   16a_1^2 + 4b_1^2 + 4b_1 + 16c_1^2 + 16c_1 + 5 = 2005
   \]
   Dividing by 4:
   \[
   4a_1^2 + b_1^2 + b_1 + 4c_1^2 + 4c_1 = 500
   \]

4. **Consider the equation modulo 4:**
   \[
   b_1^2 + b_1 \equiv 0 \pmod{4}
   \]
   This implies \(b_1 \equiv 0 \pmod{4}\) or \(b_1 \equiv -1 \pmod{4}\).

5. **Case 1: \(b_1 \equiv 0 \pmod{4}\):**
   Let \(b_1 = 4b_2\):
   \[
   4a_1^2 + 16b_2^2 + 4b_2 + 4c_1^2 + 4c_1 = 500
   \]
   Dividing by 4:
   \[
   a_1^2 + 4b_2^2 + b_2 + c_1^2 + c_1 = 125
   \]
   Since \(4b_2^2 > 125\) for \(b_2 > 5\), we have \(1 \leq b_2 \leq 5\).

   - **Subcase \(b_2 = 5\):**
     \[
     a_1^2 + c_1^2 + c_1 = 20
     \]
     Considering parity, \(c_1^2 + c_1 \equiv 0 \pmod{2}\), so \(a_1^2 = 4\) or \(16\). This yields no solutions.

   - **Subcase \(b_2 = 4\):**
     \[
     a_1^2 + c_1^2 + c_1 = 57
     \]
     Considering parity, \(a_1^2 = 1, 9, 25, 49\). This yields the solution \(c_1 = 7 \Rightarrow (a, b, c) = (4, 33, 30)\).

   - **Subcase \(b_2 = 3\):**
     \[
     a_1^2 + c_1^2 + c_1 = 86
     \]
     Considering parity, \(a_1^2 = 4, 16, 36, 64\). This yields no solutions.

   - **Subcase \(b_2 = 2\):**
     \[
     a_1^2 + c_1^2 + c_1 = 107
     \]
     Considering parity, \(a_1^2 = 1, 9, 25, 49, 81\). This yields no solutions.

   - **Subcase \(b_2 = 1\):**
     \[
     a_1^2 + c_1^2 + c_1 = 120
     \]
     Considering parity, \(a_1^2 = 4, 16, 36, 64, 100\). This yields the solutions \(c_1 = 7, 4 \Rightarrow (a, b, c) = (32, 9, 30), (40, 9, 18)\).

6. **Case 2: \(b_1 \equiv -1 \pmod{4}\):**
   Let \(b_1 = 4b_2 - 1\):
   \[
   4a_1^2 + 16b_2^2 - 4b_2 + 4c_1^2 + 4c_1 = 500
   \]
   Dividing by 4:
   \[
   a_1^2 + 4b_2^2 - b_2 + c_1^2 + c_1 = 125
   \]
   Since \(4b_2^2 - b_2 > 125\) for \(b_2 > 5\), we have \(1 \leq b_2 \leq 5\).

   - **Subcase \(b_2 = 5\):**
     \[
     a_1^2 + c_1^2 + c_1 = 30
     \]
     Considering parity, \(a_1^2 = 4, 16\). This yields no solutions.

   - **Subcase \(b_2 = 4\):**
     \[
     a_1^2 + c_1^2 + c_1 = 65
     \]
     Considering parity, \(a_1^2 = 1, 9, 25, 49\). This yields the solution \(c_1 = 7 \Rightarrow (a, b, c) = (12, 31, 30)\).

   - **Subcase \(b_2 = 3\):**
     \[
     a_1^2 + c_1^2 + c_1 = 92
     \]
     Considering parity, \(a_1^2 = 4, 16, 36, 64\). This yields the solution \(c_1 = 7 \Rightarrow (a, b, c) = (24, 23, 30)\).

   - **Subcase \(b_2 = 2\):**
     \[
     a_1^2 + c_1^2 + c_1 = 111
     \]
     Considering parity, \(a_1^2 = 1, 9, 25, 49, 81\). This yields the solutions \(c_1 = 5, 10 \Rightarrow (a, b, c) = (4, 15, 22), (36, 15, 42)\).

   - **Subcase \(b_2 = 1\):**
     \[
     a_1^2 + c_1^2 + c_1 = 122
     \]
     Considering parity, \(a_1^2 = 1, 9, 25, 49, 81, 121\). This yields no solutions.

Therefore, by considering the above cases, the only seven possible solutions are:
\[
(4, 33, 30), (32, 9, 30), (40, 9, 18), (12, 31, 30), (24, 23, 30), (4, 15, 22), (36, 15, 42)
\]

The final answer is \( \boxed{ (4, 33, 30), (32, 9, 30), (40, 9, 18), (12, 31, 30), (24, 23, 30), (4, 15, 22), (36, 15, 42) } \)