Okay, let's tackle this problem step by step. The equation given is:

\[
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
\]

First, I need to find the sum of all complex solutions to this equation. Hmm, complex solutions can include real solutions as well, but since the problem mentions complex, I should consider all possible solutions, even those that are not real. However, since the equation is a rational equation with polynomial terms on the left and a linear term on the right, maybe there's a way to simplify this.

Let me start by simplifying the left-hand side. Each term is of the form \(\frac{k}{x^2 - k}\) where k is 1, 2, 3, 4. So, maybe I can combine these fractions. Let's see:

Let me denote \( y = x^2 \). Then each term becomes \(\frac{k}{y - k}\). Maybe substituting y would make it easier to handle. Let's try that.

So substituting \( y = x^2 \), the left-hand side becomes:

\[
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}.
\]

And the right-hand side remains \(2010x - 4\). Hmm, but now we have a mix of y and x. Since \( y = x^2 \), maybe if we can express x in terms of y, but that might complicate things. Alternatively, perhaps we can combine the fractions on the left-hand side.

To combine the fractions, we can find a common denominator. The denominators are \( y - 1, y - 2, y - 3, y - 4 \). So the common denominator would be \((y - 1)(y - 2)(y - 3)(y - 4)\). Let's try combining them:

First term: \( \frac{1}{y - 1} \) times the common denominator gives \( (y - 2)(y - 3)(y - 4) \).
Second term: \( \frac{2}{y - 2} \) times the common denominator gives \( 2(y - 1)(y - 3)(y - 4) \).
Third term: \( \frac{3}{y - 3} \) times common denominator gives \( 3(y - 1)(y - 2)(y - 4) \).
Fourth term: \( \frac{4}{y - 4} \) times common denominator gives \( 4(y - 1)(y - 2)(y - 3) \).

Adding all these up, the numerator becomes:

\[
1 \cdot (y - 2)(y - 3)(y - 4) + 2 \cdot (y - 1)(y - 3)(y - 4) + 3 \cdot (y - 1)(y - 2)(y - 4) + 4 \cdot (y - 1)(y - 2)(y - 3)
\]

This seems complicated, but maybe we can expand each term and then combine like terms. Let's proceed step by step.

First term expansion:

\( (y - 2)(y - 3)(y - 4) \)

Let me compute this step by step:

First multiply (y - 2)(y - 3):

= \( y^2 - 5y + 6 \)

Then multiply by (y - 4):

= \( (y^2 - 5y + 6)(y - 4) \)
= \( y^3 - 4y^2 - 5y^2 + 20y + 6y - 24 \)
= \( y^3 - 9y^2 + 26y - 24 \)

Second term:

2*(y - 1)(y - 3)(y - 4)

First compute (y - 1)(y - 3):

= \( y^2 - 4y + 3 \)

Multiply by (y - 4):

= \( (y^2 - 4y + 3)(y - 4) \)
= \( y^3 - 4y^2 - 4y^2 + 16y + 3y - 12 \)
= \( y^3 - 8y^2 + 19y - 12 \)

Multiply by 2:

= \( 2y^3 - 16y^2 + 38y - 24 \)

Third term:

3*(y - 1)(y - 2)(y - 4)

First compute (y - 1)(y - 2):

= \( y^2 - 3y + 2 \)

Multiply by (y - 4):

= \( (y^2 - 3y + 2)(y - 4) \)
= \( y^3 - 4y^2 - 3y^2 + 12y + 2y - 8 \)
= \( y^3 - 7y^2 + 14y - 8 \)

Multiply by 3:

= \( 3y^3 - 21y^2 + 42y - 24 \)

Fourth term:

4*(y - 1)(y - 2)(y - 3)

First compute (y - 1)(y - 2):

= \( y^2 - 3y + 2 \)

Multiply by (y - 3):

= \( (y^2 - 3y + 2)(y - 3) \)
= \( y^3 - 3y^2 - 3y^2 + 9y + 2y - 6 \)
= \( y^3 - 6y^2 + 11y - 6 \)

Multiply by 4:

= \( 4y^3 - 24y^2 + 44y - 24 \)

Now, add all four expanded terms together:

First term: \( y^3 - 9y^2 + 26y - 24 \)

Second term: \( 2y^3 - 16y^2 + 38y - 24 \)

Third term: \( 3y^3 - 21y^2 + 42y - 24 \)

Fourth term: \( 4y^3 - 24y^2 + 44y - 24 \)

Adding them up:

y^3 terms: 1 + 2 + 3 + 4 = 10y^3

y^2 terms: -9 -16 -21 -24 = -70y^2

y terms: 26 + 38 + 42 + 44 = 150y

constants: -24 -24 -24 -24 = -96

So numerator is:

10y^3 -70y^2 +150y -96

Therefore, the left-hand side becomes:

\[
\frac{10y^3 -70y^2 +150y -96}{(y - 1)(y - 2)(y - 3)(y - 4)}
\]

Now, the equation is:

\[
\frac{10y^3 -70y^2 +150y -96}{(y - 1)(y - 2)(y - 3)(y - 4)} = 2010x - 4
\]

But remember that \( y = x^2 \). So substitute y back:

\[
\frac{10x^6 -70x^4 +150x^2 -96}{(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)} = 2010x - 4
\]

Wait, but the denominator when substituting y = x^2 is:

(y - 1)(y - 2)(y - 3)(y - 4) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)

So the left-hand side is a rational function in x, but the right-hand side is linear in x. Hmm, this seems complicated. Maybe cross-multiplying both sides by the denominator to eliminate the fraction. Let's try that.

Multiply both sides by (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4):

10x^6 -70x^4 +150x^2 -96 = (2010x - 4)(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)

Wow, that seems like a very high degree polynomial on the right-hand side. Let's see, expanding the right-hand side would result in a polynomial of degree 1 (from 2010x) multiplied by (x^2 - a)(x^2 - b)(x^2 - c)(x^2 - d), which is a degree 8 polynomial. So multiplying them gives a degree 9 polynomial. The left-hand side is degree 6. So the equation becomes:

10x^6 -70x^4 +150x^2 -96 = 2010x*(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) -4*(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4)

This seems extremely complicated. Maybe there's a smarter approach here. Let me think.

Alternatively, perhaps we can consider the original equation:

Sum_{k=1}^4 [k/(x^2 -k)] = 2010x -4

But maybe instead of substituting y = x^2, we can consider that each term is of the form k/(x^2 -k). Let me denote x^2 as t. Then each term becomes k/(t -k). Then the sum is sum_{k=1}^4 [k/(t -k)] = 2010x -4.

But x^2 = t, so x = sqrt(t) or x = -sqrt(t). But this might complicate things because sqrt(t) is involved, making the equation non-polynomial. Hmm.

Alternatively, maybe we can write the equation as:

Sum_{k=1}^4 [k/(x^2 -k)] = 2010x -4

Multiply both sides by the product (x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4):

Sum_{k=1}^4 [k * product_{j≠k} (x^2 -j)] = (2010x -4) * (x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4)

But this approach also leads to a high-degree polynomial. Maybe this is the way to go, but with so many terms, expanding everything seems impractical. Perhaps there's a pattern or a trick here.

Alternatively, maybe we can consider symmetry or possible roots. Let's check if x = 0 is a solution. Plugging x=0 into left-hand side:

Sum_{k=1}^4 [k/(0 -k)] = sum_{k=1}^4 [k/(-k)] = sum_{k=1}^4 (-1) = -4

Right-hand side: 2010*0 -4 = -4. So x=0 is a solution. That's one solution. So x=0 is a root. Maybe there are others.

Wait, but the problem asks for the sum of all complex solutions. So if there's a root at x=0, then the sum would be 0 plus other roots. But maybe x=0 is a repeated root? Let's check if x=0 is a multiple root.

But before that, let's check if there are other simple roots. Let's try x=1. But x=1 makes the denominators x^2 -k undefined for k=1: x^2 -1 =0, so x=1 is not allowed. Similarly for x=±1, ±sqrt(2), ±sqrt(3), ±2: these would make denominators zero, so they are excluded.

So possible solutions are other complex numbers. But since the equation is a rational equation, after cross-multiplying, we get a polynomial equation. The sum of the roots of the polynomial can be found using Vieta's formula. However, the problem is that the cross-multiplied equation is a high-degree polynomial, and Vieta's formula would require knowing all the coefficients. However, maybe we can find the degree of the polynomial and use Vieta's formula.

Let me see: the left-hand side, after cross-multiplying, is a degree 6 polynomial, as we saw earlier. The right-hand side is a degree 9 polynomial. So when we bring all terms to one side, the equation becomes:

10x^6 -70x^4 +150x^2 -96 - (2010x -4)(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4) = 0

So the highest degree term comes from expanding the right-hand side. Let's figure out the degree. The term (2010x) multiplied by (x^2)^4 (since each (x^2 -k) is degree 2, and there are four of them, so degree 8). So 2010x * x^8 = 2010x^9. Similarly, -4*(x^2)^4 = -4x^8. So the right-hand side when multiplied out will have a leading term of 2010x^9. Therefore, the left-hand side after subtracting this will have a leading term of -2010x^9. So the resulting polynomial is degree 9, with leading term -2010x^9.

But Vieta's formula tells us that the sum of the roots is equal to - (coefficient of x^8) / (coefficient of x^9). However, to find the sum, we need the coefficients of x^9 and x^8. Let's compute these coefficients.

First, the left-hand side: 10x^6 -70x^4 +150x^2 -96. The highest degree is 6, so when we subtract the right-hand side, the x^9 term comes only from the right-hand side. The right-hand side is (2010x -4)(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4). Let's compute the leading terms of this product.

First, (x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4) = (x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4). Multiplying these together, the leading term is x^2 * x^2 * x^2 * x^2 = x^8. Then multiplying by (2010x -4):

The leading term is 2010x * x^8 = 2010x^9. The next term is -4 * x^8 = -4x^8. So the right-hand side has leading terms 2010x^9 -4x^8 + ... 

Therefore, when we bring everything to the left-hand side, the equation becomes:

10x^6 -70x^4 +150x^2 -96 -2010x^9 +4x^8 - ... = 0

So the left-hand side (after moving everything to left) is:

-2010x^9 +4x^8 +10x^6 -70x^4 +150x^2 -96 - ... =0

But the dots (...) represent lower degree terms. However, the key point is that the highest degree term is -2010x^9, and the next highest degree term is +4x^8. Therefore, the polynomial is of degree 9, with leading coefficient -2010 and the coefficient of x^8 is +4. 

Therefore, by Vieta's formula, the sum of all roots is equal to - (coefficient of x^8) / (coefficient of x^9). So that's - (4)/(-2010) = 4/2010 = 2/1005.

Wait, but the problem is asking for the sum of all complex solutions. However, the equation after cross-multiplying includes the original equation's denominators, which are x^2 -1, x^2 -2, etc. Each of these would exclude x = ±1, ±√2, ±√3, ±2. So the original equation has some restrictions on x. However, when we cross-multiply, we might introduce extraneous roots, but the problem states "sum of all complex solutions" of the original equation. So perhaps these excluded x values are not roots of the original equation, but could be roots of the cross-multiplied equation. Therefore, we need to make sure that when we find the roots of the cross-multiplied polynomial, we exclude the extraneous roots introduced by the cross-multiplication.

But this complicates things. However, perhaps the original equation is equivalent to the cross-multiplied equation except for the excluded roots. Therefore, the sum of the roots of the original equation would be equal to the sum of the roots of the cross-multiplied polynomial minus the sum of the excluded roots. But the excluded roots are x = ±1, ±√2, ±√3, ±2, which are 8 roots. However, in the cross-multiplied polynomial, these excluded roots would have made the denominator zero, so they are not roots of the cross-multiplied polynomial. Therefore, the cross-multiplied polynomial does not have these as roots, so the roots of the original equation are exactly the roots of the cross-multiplied polynomial. Therefore, the sum of the roots of the original equation is the same as the sum of the roots of the cross-multiplied polynomial, which is 2/1005.

But wait, hold on. The cross-multiplied polynomial is of degree 9. The original equation is equivalent to the cross-multiplied equation, except for the excluded roots. However, the cross-multiplied equation includes the original equation's solutions and potentially other roots that were excluded. Wait, no. If we cross-multiply, the equation becomes:

Left-hand side * denominator = right-hand side * denominator

So if the original equation is LHS = RHS, then cross-multiplying gives LHS * denominator = RHS * denominator, which is equivalent to (LHS - RHS) * denominator = 0. Therefore, the roots are the roots of LHS - RHS =0, plus the roots where denominator =0. However, the original equation excludes the roots where denominator=0. Therefore, the cross-multiplied equation has roots at the solutions of the original equation plus the roots where denominator=0. Therefore, if we take the cross-multiplied equation, the roots are the solutions to the original equation plus the excluded roots. Therefore, the sum of all roots of the cross-multiplied equation is the sum of the solutions to the original equation plus the sum of the excluded roots. But the sum of the excluded roots is the sum of x = ±1, ±√2, ±√3, ±2. Let's compute that:

Sum = 1 + (-1) + √2 + (-√2) + √3 + (-√3) + 2 + (-2) = 0.

So the sum of the excluded roots is zero. Therefore, the sum of the roots of the cross-multiplied polynomial is equal to the sum of the solutions of the original equation. Therefore, the sum of the complex solutions of the original equation is equal to the sum of all roots of the cross-multiplied polynomial, which is 2/1005.

Therefore, the answer is 2/1005. However, let me verify this again.

Wait, Vieta's formula gives that the sum of all roots (including the excluded ones) is 2/1005. Since the sum of the excluded roots is zero, the sum of the original equation's solutions is also 2/1005. Therefore, the answer is 2/1005. So boxed, that would be \boxed{\dfrac{2}{1005}}.

But let me check if there's a miscalculation in the coefficients. Let's see:

The cross-multiplied polynomial is:

-2010x^9 +4x^8 +10x^6 -70x^4 +150x^2 -96 - ... =0

Wait, but the left-hand side after moving everything to the left is:

10x^6 -70x^4 +150x^2 -96 - [2010x*(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4) -4*(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4)] =0

So expanding the right-hand side:

2010x*(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4) -4*(x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4)

So when we subtract this from the left-hand side, we get:

Left-hand side - (2010x*(product) -4*(product)) =0

Which is equivalent to:

Left-hand side -2010x*(product) +4*(product) =0

Therefore, the equation is:

(10x^6 -70x^4 +150x^2 -96) + ( -2010x*(product) +4*(product)) =0

But when considering the highest degree terms, the product term is (x^2 -1)(x^2 -2)(x^2 -3)(x^2 -4) which is degree 8. Multiplying by x gives degree 9. So the term -2010x*(product) has degree 9, and +4*(product) has degree 8. The left-hand side has degree 6. Therefore, when combining, the highest degree term is -2010x^9, next is +4x^8, then the rest. So the leading coefficients are -2010 and +4, so the sum of roots is - (4)/(-2010) = 4/2010 = 2/1005.

Yes, that seems correct. So the sum of all complex solutions is 2/1005.

**Final Answer**
\boxed{\dfrac{2}{1005}}
Therefore, the sum of all complex solutions of the given equation is $\boxed{\frac{2}{1005}}$.
</think>