Given the equalities \(\frac{a}{2003} = \frac{b}{2004} = \frac{c}{2005} = k\), we can express \(a\), \(b\), and \(c\) in terms of \(k\):

\[
a = 2003k, \quad b = 2004k, \quad c = 2005k.
\]

Next, we calculate the differences \(a-b\), \(b-c\), and \(c-a\):

1. **Calculate \(a - b\)**:
\[
a - b = 2003k - 2004k = -k.
\]

2. **Calculate \(b - c\)**:
\[
b - c = 2004k - 2005k = -k.
\]

3. **Calculate \(c - a\)**:
\[
c - a = 2005k - 2003k = 2k.
\]

Now, we substitute these values into the equation we want to prove, \(4(a-b)(b-c) = (c-a)^2\):

**Left Side**:
\[
4(a-b)(b-c) = 4(-k)(-k) = 4k^2.
\]

**Right Side**:
\[
(c-a)^2 = (2k)^2 = 4k^2.
\]

Since both sides are equal:
\[
4(a-b)(b-c) = (c-a)^2,
\]
we have shown that the statement is true. Thus, the proof is completed:

\[
\boxed{4(a-b)(b-c) = (c-a)^2}.
\]