Find all odd postive integers less than $2007$ such that the sum of all of its positive divisors is odd.
1. **Express \( n \) in terms of its prime factors:**
   Let \( n = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k} \) be an odd positive integer. Since \( n \) is odd, none of the prime factors \( p_i \) are equal to 2.

2. **Sum of divisors formula:**
   The sum of the divisors of \( n \) is given by:
   \[
   \sigma(n) = (1 + p_1 + p_1^2 + \ldots + p_1^{e_1})(1 + p_2 + p_2^2 + \ldots + p_2^{e_2}) \ldots (1 + p_k + \ldots + p_k^{e_k})
   \]
   We need this sum to be odd.

3. **Oddness of each term in the product:**
   For the product to be odd, each term \( (1 + p_i + p_i^2 + \ldots + p_i^{e_i}) \) must be odd. Since \( p_i \) is an odd prime, each \( p_i^j \) is also odd. Therefore, the sum \( 1 + p_i + p_i^2 + \ldots + p_i^{e_i} \) is a sum of odd numbers.

4. **Modulo 2 analysis:**
   Consider the sum modulo 2:
   \[
   1 + p_i + p_i^2 + \ldots + p_i^{e_i} \equiv 1 + 1 + 1 + \ldots + 1 \equiv e_i + 1 \pmod{2}
   \]
   For this sum to be odd, \( e_i + 1 \) must be odd, which implies \( e_i \) must be even.

5. **Conclusion about \( n \):**
   Since \( e_i \) is even for all \( i \), \( n \) must be a perfect square (as each exponent in the prime factorization is even).

6. **Verification:**
   Conversely, if \( n \) is an odd perfect square, then each \( e_i \) is even, and the sum of the divisors \( \sigma(n) \) will be odd. Therefore, all odd perfect squares less than 2007 satisfy the condition.

7. **List of odd perfect squares less than 2007:**
   The odd perfect squares less than 2007 are:
   \[
   1^2, 3^2, 5^2, 7^2, 9^2, 11^2, 13^2, 15^2, 17^2, 19^2, 21^2, 23^2, 25^2, 27^2, 29^2, 31^2, 33^2, 35^2, 37^2, 39^2, 41^2, 43^2
   \]
   Calculating these:
   \[
   1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849
   \]

The final answer is \(\boxed{1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849}\)