Let $n\geq 2$ be an integer and let $f$ be a $4n$-variable polynomial with real coefficients. Assume that, for any $2n$ points $(x_1,y_1),\dots,(x_{2n},y_{2n})$ in the Cartesian plane, $f(x_1,y_1,\dots,x_{2n},y_{2n})=0$ if and only if the points form the vertices of a regular $2n$-gon in some order, or are all equal. 

Determine the smallest possible degree of $f$.

(Note, for example, that the degree of the polynomial $$g(x,y)=4x^3y^4+yx+x-2$$ is $7$ because $7=3+4$.)

[i]Ankan Bhattacharya[/i]
To determine the smallest possible degree of the polynomial \( f \), we need to analyze the conditions under which \( f \) evaluates to zero. The polynomial \( f \) is zero if and only if the given \( 2n \) points form the vertices of a regular \( 2n \)-gon or are all equal. 

### Step 1: Establishing the Sign of \( f \)
First, we note a key property of \( f \):

**Claim:** Either \( f \ge 0 \) or \( f \le 0 \).

**Proof:** This follows from the fact that the zero-set of \( f \) is very sparse. If \( f \) takes on both positive and negative values, we can move the points \( A_1, \dots, A_{2n} \) from a configuration where \( f \) is negative to one where \( f \) is positive without ever having them form a regular \( 2n \)-gon, which contradicts the given condition. Therefore, \( f \) must be non-negative or non-positive. \(\blacksquare\)

### Step 2: Degree Bound
To show that \(\deg f \ge 2n\), we will animate the points \( A_1, \dots, A_{2n} \) linearly in a variable \( t \). Define \( g(t) = f(A_1(t), \dots, A_{2n}(t)) \). The degree of \( g(t) \) is at most \(\deg f\), assuming \( g(t) \) is not identically zero.

### Step 3: Multiple Roots
The previous lemma establishes that any root of \( g(t) \) must be a multiple root. If we can show that there are at least \( n \) roots, we will have shown \(\deg g \ge 2n\), and thus \(\deg f \ge 2n\).

### Step 4: Geometric Construction
We need to exhibit \( 2n \) linearly moving points such that they form a regular \( 2n \)-gon a total of \( n \) times, but not always form one. 

Draw \( n \) mirrors through the origin, as lines making angles of \(\frac{\pi}{n}\) with each other. Any point \( P \) has a total of \( 2n \) reflections in the mirrors. Draw the \( n \) angle bisectors of adjacent mirrors. The reflections of \( P \) form a regular \( 2n \)-gon if and only if \( P \) lies on one of the bisectors.

### Step 5: Animation
Animate \( P \) on any line which intersects all \( n \) bisectors, and let \( P_1, \dots, P_{2n} \) be its reflections. These points will form a regular \( 2n \)-gon at least \( n \) times. They will not always form a regular \( 2n \)-gon, establishing \(\deg f \ge 2n\).

### Step 6: Construction of \( f \)
We will construct \( f \) as a sum of squares. Write \( f = f_1^2 + \dots + f_k^2 \). If their degrees are \( d_1, \dots, d_k \), then \( f \) has degree at most \( 2 \max(d_1, \dots, d_k) \).

### Step 7: Constraints
We need polynomials of degree at most \( n \) such that \( 2n \) points are the vertices of a regular \( 2n \)-gon if and only if the polynomials are all zero at those points. Assume \( A_1 + \dots + A_{2n} = 0 \). 

First, impose the constraints that all \( |A_i|^2 = x_i^2 + y_i^2 \) are equal. This uses multiple degree 2 constraints. Now, the points lie on a circle with center 0. If the circle has radius 0, all \( A_i \) coincide.

### Step 8: Lemma for Regular \( 2n \)-gon
**Lemma:** Suppose \( a_1, \dots, a_{2n} \) are complex numbers of the same nonzero magnitude, and \( a_1^k + \dots + a_{2n}^k = 0 \) for all \( k \in \{1, \dots, n\} \). Then \( a_1, \dots, a_{2n} \) form a regular \( 2n \)-gon centered at the origin.

**Proof:** Assume \( a_1, \dots, a_{2n} \) lie on the unit circle. By Newton's sums, the \( k \)-th symmetric sums of \( a_1, \dots, a_{2n} \) are zero for \( k \in \{1, \dots, n\} \). Taking conjugates, \( a_1^{-k} + \dots + a_{2n}^{-k} = 0 \) for \( k \in \{1, \dots, n\} \). The first \( 2n-1 \) symmetric sums of \( a_1, \dots, a_{2n} \) are zero, implying they form a regular \( 2n \)-gon. \(\blacksquare\)

### Step 9: Encoding Constraints
Write \( a_r = x_r + y_r i \). The constraint \( a_1^k + \dots + a_{2n}^k = 0 \) can be expressed as \( p_k + q_k i = 0 \), where \( p_k \) and \( q_k \) are real polynomials. Impose \( p_k = 0 \) and \( q_k = 0 \), conditions of degree \( k \le n \).

### Step 10: Final Polynomial
Taking the sum of squares of all these constraints gives a polynomial \( f \) of degree at most \( 2n \) which works whenever \( A_1 + \dots + A_{2n} = 0 \). The centroid-shifting trick gives a polynomial which works in general.

\(\blacksquare\)

The final answer is \( \boxed{ 2n } \)