Let $n$ be a positive integer, and let $a>0$ be a real number. Consider the equation: 
\[ \sum_{i=1}^{n}(x_i^2+(a-x_i)^2)= na^2 \]
How many solutions ($x_1, x_2 \cdots , x_n$) does this equation have, such that:
\[ 0 \leq x_i \leq a, i \in N^+ \]
1. Consider the given equation:
   \[
   \sum_{i=1}^{n}(x_i^2+(a-x_i)^2)= na^2
   \]
   We need to find the number of solutions \((x_1, x_2, \ldots, x_n)\) such that \(0 \leq x_i \leq a\) for all \(i\).

2. First, simplify the expression inside the summation:
   \[
   x_i^2 + (a - x_i)^2 = x_i^2 + a^2 - 2ax_i + x_i^2 = 2x_i^2 - 2ax_i + a^2
   \]
   Therefore, the equation becomes:
   \[
   \sum_{i=1}^{n} (2x_i^2 - 2ax_i + a^2) = na^2
   \]

3. Distribute the summation:
   \[
   2 \sum_{i=1}^{n} x_i^2 - 2a \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} a^2 = na^2
   \]
   Since \(\sum_{i=1}^{n} a^2 = na^2\), the equation simplifies to:
   \[
   2 \sum_{i=1}^{n} x_i^2 - 2a \sum_{i=1}^{n} x_i + na^2 = na^2
   \]

4. Subtract \(na^2\) from both sides:
   \[
   2 \sum_{i=1}^{n} x_i^2 - 2a \sum_{i=1}^{n} x_i = 0
   \]

5. Factor out the common term of 2:
   \[
   \sum_{i=1}^{n} x_i^2 - a \sum_{i=1}^{n} x_i = 0
   \]

6. Rearrange the equation:
   \[
   \sum_{i=1}^{n} x_i^2 = a \sum_{i=1}^{n} x_i
   \]

7. Let \(S = \sum_{i=1}^{n} x_i\). Then the equation becomes:
   \[
   \sum_{i=1}^{n} x_i^2 = aS
   \]

8. Notice that \(x_i^2 \leq ax_i\) for \(0 \leq x_i \leq a\). Equality holds if and only if \(x_i = 0\) or \(x_i = a\).

9. Therefore, for the equation \(\sum_{i=1}^{n} x_i^2 = aS\) to hold, each \(x_i\) must be either 0 or \(a\).

10. Each \(x_i\) has 2 choices: 0 or \(a\). Since there are \(n\) such \(x_i\), the total number of solutions is:
    \[
    2^n
    \]

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