Find all triples $(a, b, c)$ of positive integers for which $$\begin{cases} a + bc=2010 \\ b + ca = 250\end{cases}$$
1. We start with the given system of equations:
   \[
   \begin{cases}
   a + bc = 2010 \\
   b + ca = 250
   \end{cases}
   \]

2. Summing both equations, we get:
   \[
   a + bc + b + ca = 2010 + 250
   \]
   Simplifying, we have:
   \[
   a + b + bc + ca = 2260
   \]
   Factoring, we get:
   \[
   (a + b)(1 + c) = 2260
   \]
   Therefore, \(1 + c\) must be a divisor of 2260. The divisors of 2260 are:
   \[
   1, 2, 4, 5, 10, 20, 113, 226, 453, 565, 1130, 2260
   \]
   Hence, the possible values for \(1 + c\) are:
   \[
   2, 4, 5, 10, 20, 113, 226, 453, 565, 1130, 2260
   \]
   This implies the possible values for \(c\) are:
   \[
   1, 3, 4, 9, 19, 112, 225, 452, 564, 1129, 2259
   \]

3. Next, we subtract the second equation from the first:
   \[
   a + bc - (b + ca) = 2010 - 250
   \]
   Simplifying, we get:
   \[
   a - b + bc - ca = 1760
   \]
   Factoring, we have:
   \[
   (b - a)(c - 1) = 1760
   \]
   Therefore, \(c - 1\) must be a divisor of 1760. The divisors of 1760 are:
   \[
   1, 2, 4, 5, 8, 10, 11, 16, 20, 22, 32, 40, 44, 55, 80, 88, 110, 160, 176, 220, 352, 440, 880, 1760
   \]
   Hence, the possible values for \(c - 1\) are:
   \[
   1, 2, 4, 5, 8, 10, 11, 16, 20, 22, 32, 40, 44, 55, 80, 88, 110, 160, 176, 220, 352, 440, 880, 1760
   \]
   This implies the possible values for \(c\) are:
   \[
   2, 3, 5, 6, 9, 11, 12, 17, 21, 23, 33, 41, 45, 56, 81, 89, 111, 161, 177, 221, 353, 441, 881, 1761
   \]

4. We now look for common values of \(c\) from both sets of possible values:
   \[
   1, 3, 4, 9, 19, 112, 225, 452, 564, 1129, 2259
   \]
   and
   \[
   2, 3, 5, 6, 9, 11, 12, 17, 21, 23, 33, 41, 45, 56, 81, 89, 111, 161, 177, 221, 353, 441, 881, 1761
   \]
   The common values are:
   \[
   3, 9
   \]

5. We test \(c = 9\):
   \[
   (a + b)(1 + 9) = 2260 \implies (a + b) \cdot 10 = 2260 \implies a + b = 226
   \]
   \[
   (b - a)(9 - 1) = 1760 \implies (b - a) \cdot 8 = 1760 \implies b - a = 220
   \]

6. Solving the system:
   \[
   \begin{cases}
   a + b = 226 \\
   b - a = 220
   \end{cases}
   \]
   Adding these equations, we get:
   \[
   2b = 446 \implies b = 223
   \]
   Subtracting the second from the first, we get:
   \[
   2a = 6 \implies a = 3
   \]

7. Therefore, the solution is:
   \[
   (a, b, c) = (3, 223, 9)
   \]

The final answer is \( \boxed{ (3, 223, 9) } \)