Can we find positive reals $a_1, a_2, \dots, a_{2002}$ such that for any positive integer $k$, with $1 \leq k \leq 2002$, every complex root $z$ of the following polynomial $f(x)$ satisfies the condition $|\text{Im } z| \leq |\text{Re } z|$, 
    \[f(x)=a_{k+2001}x^{2001}+a_{k+2000}x^{2000}+ \cdots + a_{k+1}x+a_k,\]  where $a_{2002+i}=a_i$, for $i=1,2, \dots, 2001$.
To determine whether there exist positive reals \( a_1, a_2, \dots, a_{2002} \) such that for any positive integer \( k \) with \( 1 \leq k \leq 2002 \), every complex root \( z \) of the polynomial 
\[ f(x) = a_{k+2001}x^{2001} + a_{k+2000}x^{2000} + \cdots + a_{k+1}x + a_k \]
satisfies the condition \( |\text{Im } z| \leq |\text{Re } z| \), we need to analyze the given polynomial and the condition on its roots.

1. **Understanding the Condition on Roots:**
   The condition \( |\text{Im } z| \leq |\text{Re } z| \) implies that the imaginary part of any root \( z \) is at most the absolute value of its real part. This means that the roots lie within the region bounded by the lines \( y = x \) and \( y = -x \) in the complex plane.

2. **Analyzing the Polynomial:**
   The polynomial \( f(x) \) is given by:
   \[ f(x) = a_{k+2001}x^{2001} + a_{k+2000}x^{2000} + \cdots + a_{k+1}x + a_k, \]
   where \( a_{2002+i} = a_i \) for \( i = 1, 2, \dots, 2001 \). This means the coefficients are periodic with period 2002.

3. **Roots of Unity:**
   Consider the polynomial \( P(x) = (x - \omega)(x - \omega^2) \cdots (x - \omega^{2002}) \), where \( \omega = e^{2\pi i / 2002} \) is a primitive 2002-th root of unity. The roots of this polynomial are evenly distributed on the unit circle in the complex plane.

4. **Condition on Roots:**
   For the roots of \( f(x) \) to satisfy \( |\text{Im } z| \leq |\text{Re } z| \), they must lie within the specified region. However, the roots of \( P(x) \) are symmetrically distributed around the unit circle, and not all of them will satisfy \( |\text{Im } z| \leq |\text{Re } z| \).

5. **Constructing the Polynomial:**
   To construct a polynomial whose roots satisfy the given condition, we need to ensure that the roots lie within the specified region. One way to achieve this is to consider polynomials with real coefficients, as their roots will be symmetric with respect to the real axis.

6. **Real Coefficients:**
   If \( f(x) \) has real coefficients, then for any complex root \( z \), its conjugate \( \overline{z} \) is also a root. This symmetry ensures that the imaginary parts of the roots are balanced, and the condition \( |\text{Im } z| \leq |\text{Re } z| \) can be satisfied.

7. **Conclusion:**
   Given the periodic nature of the coefficients and the requirement for the roots to lie within a specific region, it is possible to choose positive reals \( a_1, a_2, \dots, a_{2002} \) such that the polynomial \( f(x) \) has roots satisfying \( |\text{Im } z| \leq |\text{Re } z| \). One possible choice is to ensure that the coefficients are chosen such that the polynomial has real coefficients, which guarantees the desired root condition.

The final answer is \( \boxed{ a_1, a_2, \dots, a_{2002} } \) can be chosen such that the polynomial \( f(x) \) satisfies the given condition on its roots.