Find all polynomials $P(x)$ with real coefficients such that

$$
(x-2010) P(x+67)=x P(x)
$$

for every integer $x$.
Taking \( x=0 \) in the given equality leads to \( -2010 P(67)=0 \), implying \( P(67)=0 \). Whenever \( i \) is an integer such that \( 1 \leq i < 30 \) and \( P(i \cdot 67)=0 \), taking \( x=i \cdot 67 \) leads to \( (i \cdot 67 - 2010) P((i+1) \cdot 67)=0 \); as \( i \cdot 67 < 2010 \) for \( i < 30 \), this implies \( P((i+1) \cdot 67)=0 \). Thus, by induction, \( P(i \cdot 67)=0 \) for all \( i=1,2, \ldots, 30 \). Hence

\[
P(x) \equiv (x-67)(x-2 \cdot 67) \ldots (x-30 \cdot 67) Q(x)
\]

where \( Q(x) \) is another polynomial.

Substituting this expression for \( P \) in the original equality, one obtains

\[
(x-2010) \cdot x(x-67) \ldots (x-29 \cdot 67) Q(x+67) = x(x-67)(x-2 \cdot 67) \ldots (x-30 \cdot 67) Q(x)
\]

which is equivalent to

\[
x(x-67)(x-2 \cdot 67) \ldots (x-30 \cdot 67)(Q(x+67) - Q(x)) = 0.
\]

By conditions of the problem, this holds for every integer \( x \). Hence there are infinitely many roots of polynomial \( Q(x+67) - Q(x) \), implying that \( Q(x+67) - Q(x) \equiv 0 \). Let \( c = Q(0) \); then \( Q(i \cdot 67) = c \) for every integer \( i \) by easy induction. Thus polynomial \( Q(x) - c \) has infinitely many roots whence \( Q(x) \equiv c \).

Consequently, \( P(x) = c(x-67)(x-2 \cdot 67) \ldots (x-30 \cdot 67) \) for some real number \( c \). As equation (1) shows, all such polynomials fit.