Let $S_n$ be number of ordered sets of natural numbers $(a_1;a_2;....;a_n)$ for which $\frac{1}{a_1}+\frac{1}{a_2}+....+\frac{1}{a_n}=1$. Determine 
1)$S_{10} mod(2)$.
2)$S_7 mod(2)$.

(1) is first problem in 10 grade, (2)- third in 9 grade.
### Part (a): Determine \( S_{10} \mod 2 \)

1. **Identify the problem constraints**:
   We need to find the number of ordered sets of natural numbers \((a_1, a_2, \ldots, a_{10})\) such that:
   \[
   \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_{10}} = 1
   \]
   and determine \( S_{10} \mod 2 \).

2. **Consider the permutations**:
   Since the set is ordered, for any satisfying (multi)set \(\{a_1, a_2, \cdots, a_{10}\}\), the number of permutations of the set is divisible by 2 unless all elements are equal or the set is in the pattern \(\{a,a,a,a,a,a,a,a,b,b\}\).

3. **Case 1: All elements are equal**:
   The only set where all elements are equal is \(\{10, 10, \cdots, 10\}\). This set contributes exactly 1 permutation.

4. **Case 2: Pattern \(\{a,a,a,a,a,a,a,a,b,b\}\)**:
   We need to solve:
   \[
   \frac{8}{a_1} + \frac{2}{a_9} = 1 \implies 8a_9 + 2a_1 = a_1a_9 \implies (a_1 - 8)(a_9 - 2) = 16
   \]
   The solutions to this equation are:
   \[
   (a_1, a_9) = (9, 18), (12, 6), (16, 4), (24, 3)
   \]
   This gives the sets:
   \[
   \{9,9,9,9,9,9,9,9,18,18\}, \{12,12,12,12,12,12,12,12,6,6\}, \{16,16,16,16,16,16,16,16,4,4\}, \{24,24,24,24,24,24,24,24,3,3\}
   \]
   Each of these sets contributes 45 permutations, and since \(45 \equiv 1 \mod 2\), they do not affect the modulo.

5. **Conclusion for \( S_{10} \mod 2 \)**:
   The only set that affects the modulo is \(\{10, 10, \cdots, 10\}\), which contributes 1 permutation. Therefore:
   \[
   S_{10} \mod 2 = 1
   \]

### Part (b): Determine \( S_7 \mod 2 \)

1. **Identify the problem constraints**:
   We need to find the number of ordered sets of natural numbers \((a_1, a_2, \ldots, a_7)\) such that:
   \[
   \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_7} = 1
   \]
   and determine \( S_7 \mod 2 \).

2. **Consider the permutations**:
   Similar to part (a), we consider the cases where the number of permutations is 1 mod 2:
   - All elements are equal.
   - 6 elements are equal, the other one is different.
   - 5 elements are equal, the other two are equal but different from the five.

3. **Case 1: All elements are equal**:
   The only set where all elements are equal is \(\{7, 7, \cdots, 7\}\). This set contributes exactly 1 permutation.

4. **Case 2: Pattern \(\{a,a,a,a,a,a,b\}\)**:
   We need to solve:
   \[
   \frac{6}{a_1} + \frac{1}{a_7} = 1 \implies 6a_7 + a_1 = a_1a_7 \implies (a_1 - 6)(a_7 - 1) = 6
   \]
   The solutions to this equation are:
   \[
   (a_1, a_7) = (8, 4), (9, 3), (12, 2)
   \]
   This gives the sets:
   \[
   \{8,8,8,8,8,8,4\}, \{9,9,9,9,9,9,3\}, \{12,12,12,12,12,12,2\}
   \]
   Each of these sets contributes 7 permutations, and since \(7 \equiv 1 \mod 2\), they contribute 21 permutations in total.

5. **Case 3: Pattern \(\{a,a,a,a,a,b,b\}\)**:
   We need to solve:
   \[
   \frac{5}{a_1} + \frac{2}{a_6} = 1 \implies 5a_6 + 2a_1 = a_1a_6 \implies (a_1 - 5)(a_6 - 2) = 10
   \]
   The solutions to this equation are:
   \[
   (a_1, a_6) = (6, 12), (10, 4), (15, 3)
   \]
   This gives the sets:
   \[
   \{6,6,6,6,6,12,12\}, \{10,10,10,10,10,4,4\}, \{15,15,15,15,15,3,3\}
   \]
   Each of these sets contributes 21 permutations, and since \(21 \equiv 1 \mod 2\), they contribute 63 permutations in total.

6. **Conclusion for \( S_7 \mod 2 \)**:
   Summing up the contributions:
   \[
   1 + 21 + 63 = 85 \equiv 1 \mod 2
   \]
   Therefore:
   \[
   S_7 \mod 2 = 1
   \]

The final answer is \( \boxed{ S_{10} \mod 2 = 1 } \) and \( S_7 \mod 2 = 1 \)