Given $100$ quadratic polynomials $f_1(x)=ax^2+bx+c_1, ... f_{100}(x)=ax^2+bx+c_{100}$. One selected $x_1, x_2... x_{100}$ - roots of $f_1, f_2, ... f_{100}$ respectively.What is the value of sum $f_2(x_1)+...+f_{100}(x_{99})+f_1(x_{100})?$
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Also 9.1 in 3rd round of Russian National Olympiad
1. We start with the given quadratic polynomials:
   \[
   f_1(x) = ax^2 + bx + c_1, \quad f_2(x) = ax^2 + bx + c_2, \quad \ldots, \quad f_{100}(x) = ax^2 + bx + c_{100}
   \]
   and the roots \(x_1, x_2, \ldots, x_{100}\) such that \(f_1(x_1) = 0\), \(f_2(x_2) = 0\), \(\ldots\), \(f_{100}(x_{100}) = 0\).

2. We need to find the value of the sum:
   \[
   f_2(x_1) + f_3(x_2) + \ldots + f_{100}(x_{99}) + f_1(x_{100})
   \]

3. Notice that for each \(i\), \(f_i(x_i) = 0\). Therefore, we can express each polynomial \(f_i(x)\) in terms of \(f_{i-1}(x)\):
   \[
   f_{i}(x) = ax^2 + bx + c_i
   \]
   Since \(f_{i-1}(x) = ax^2 + bx + c_{i-1}\), we have:
   \[
   f_{i}(x) = f_{i-1}(x) + c_i - c_{i-1}
   \]

4. Using this relationship, we can write:
   \[
   f_2(x_1) = f_1(x_1) + c_2 - c_1
   \]
   Since \(f_1(x_1) = 0\), it follows that:
   \[
   f_2(x_1) = c_2 - c_1
   \]

5. Similarly, for the other terms:
   \[
   f_3(x_2) = f_2(x_2) + c_3 - c_2 = c_3 - c_2
   \]
   \[
   f_4(x_3) = f_3(x_3) + c_4 - c_3 = c_4 - c_3
   \]
   \[
   \vdots
   \]
   \[
   f_{100}(x_{99}) = f_{99}(x_{99}) + c_{100} - c_{99} = c_{100} - c_{99}
   \]
   \[
   f_1(x_{100}) = f_{100}(x_{100}) + c_1 - c_{100} = c_1 - c_{100}
   \]

6. Summing all these terms, we get:
   \[
   f_2(x_1) + f_3(x_2) + \ldots + f_{100}(x_{99}) + f_1(x_{100}) = (c_2 - c_1) + (c_3 - c_2) + \ldots + (c_{100} - c_{99}) + (c_1 - c_{100})
   \]

7. Notice that in the sum, each \(c_i\) cancels out:
   \[
   (c_2 - c_1) + (c_3 - c_2) + \ldots + (c_{100} - c_{99}) + (c_1 - c_{100}) = 0
   \]

Thus, the value of the sum is \(0\).

The final answer is \(\boxed{0}\)