Let $n$ be a positive integer. $2n+1$ tokens are in a row, each being black or white. A token is said to be [i]balanced[/i] if the number of white tokens on its left plus the number of black tokens on its right is $n$. Determine whether the number of [i]balanced[/i] tokens is even or odd.
1. **Initial Configuration Analysis**:
   - Consider an initial configuration where the first \( n+1 \) tokens are black and the remaining \( n \) tokens are white. This configuration looks like:
     \[
     \underbrace{BBBB \cdots B}_{n+1} \underbrace{WWWW \cdots W}_{n}
     \]
   - We need to determine the number of balanced tokens in this configuration. A token is balanced if the number of white tokens on its left plus the number of black tokens on its right is \( n \).

2. **Counting Balanced Tokens in Initial Configuration**:
   - For the first \( n+1 \) black tokens, there are no white tokens on their left. Therefore, none of these tokens can be balanced because the number of white tokens on their left is zero, which is less than \( n \).
   - For the last \( n \) white tokens, there are \( n+1 \) black tokens on their left. Therefore, the number of white tokens on their left plus the number of black tokens on their right for each of these white tokens is:
     \[
     n+1 + 0 = n+1
     \]
     which is greater than \( n \). Hence, none of these white tokens can be balanced either.
   - The only token that can be balanced is the token at position \( n+1 \) (the last black token in the initial configuration). For this token, the number of white tokens on its left is \( n \) and the number of black tokens on its right is zero:
     \[
     n + 0 = n
     \]
     Therefore, this token is balanced.

3. **Parity Invariance Under Token Swaps**:
   - Consider the operation of swapping a black token \( B \) with a white token \( W \) such that \( \cdots BW \cdots \rightarrow \cdots WB \cdots \).
   - We need to analyze how this swap affects the balanced status of the tokens:
     - If \( B \) was balanced before the swap, then \( W \) will be balanced after the swap, and vice versa.
     - If both \( B \) and \( W \) were balanced before the swap, they will both become unbalanced after the swap.
     - If both \( B \) and \( W \) were unbalanced before the swap, they will either both remain unbalanced or both become balanced after the swap.
   - Therefore, the parity (odd or even nature) of the number of balanced tokens remains unchanged under any such swap.

4. **Conclusion**:
   - Since the initial configuration has exactly one balanced token (which is odd), and the parity of the number of balanced tokens remains invariant under any sequence of swaps, the number of balanced tokens in any configuration must always be odd.

\(\blacksquare\)

The final answer is \( \boxed{ \text{Odd} } \)