Find all the odd positive integers $n$ such that there are $n$ odd integers $x_1, x_2,..., x_n$ such that
$$x_1^2+x_2^2+...+x_n^2=n^4$$
1. **Claim**: We claim that \( n \) has the desired property if and only if \( n \equiv 1 \pmod{8} \).

2. **Necessity**: First, we need to show that if \( n \) has the desired property, then \( n \equiv 1 \pmod{8} \).

   - Consider the equation \( x_1^2 + x_2^2 + \cdots + x_n^2 = n^4 \).
   - Since \( x_i \) are odd integers, \( x_i^2 \equiv 1 \pmod{8} \) for each \( i \).
   - Therefore, \( x_1^2 + x_2^2 + \cdots + x_n^2 \equiv n \pmod{8} \).
   - On the other hand, \( n^4 \equiv 1 \pmod{8} \) because \( n \) is odd and \( n^4 \equiv 1^4 \equiv 1 \pmod{8} \).
   - Hence, \( n \equiv 1 \pmod{8} \).

3. **Sufficiency**: Next, we need to show that if \( n \equiv 1 \pmod{8} \), then there exist \( n \) odd integers \( x_1, x_2, \ldots, x_n \) such that \( x_1^2 + x_2^2 + \cdots + x_n^2 = n^4 \).

   - Assume \( n = 8k + 1 \) for some natural number \( k \).
   - We need to construct \( n \) odd integers \( x_1, x_2, \ldots, x_n \) such that their squares sum to \( n^4 \).
   - Consider the following construction:
     \[
     \begin{aligned}
     &a_1 = n^2 - 2, \\
     &a_2 = 2n - 1, \\
     &a_3 = a_4 = \cdots = a_{3k+2} = 3, \\
     &a_{3k+3} = a_{3k+4} = \cdots = a_{8k+1} = 1.
     \end{aligned}
     \]
   - We need to verify that the sum of the squares of these integers equals \( n^4 \):
     \[
     \begin{aligned}
     &a_1^2 + a_2^2 + \sum_{i=3}^{3k+2} a_i^2 + \sum_{i=3k+3}^{8k+1} a_i^2 \\
     &= (n^2 - 2)^2 + (2n - 1)^2 + (3k \cdot 3^2) + ((5k + 1) \cdot 1^2) \\
     &= (n^2 - 2)^2 + (2n - 1)^2 + 9k + (5k + 1) \\
     &= n^4 - 4n^2 + 4 + 4n^2 - 4n + 1 + 9k + 5k + 1 \\
     &= n^4 + 6k + 6k + 2 \\
     &= n^4.
     \end{aligned}
     \]
   - This construction works, and thus \( n \equiv 1 \pmod{8} \) is sufficient.

Therefore, the proof is complete.

The final answer is \( \boxed{ n \equiv 1 \pmod{8} } \).