Given a positive integer $ n$, find all integers $ (a_{1},a_{2},\cdots,a_{n})$ satisfying the following conditions:
$ (1): a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{n}\ge n^2;$
$ (2): a_{1}^2\plus{}a_{2}^2\plus{}\cdots\plus{}a_{n}^2\le n^3\plus{}1.$
1. **Assume the order of the sequence**: Without loss of generality, let \( a_n \ge a_{n-1} \ge \cdots \ge a_1 \).

2. **Define sums**: Let \( \sum_{i=1}^{n-1} a_i = A_1 \) and \( \sum_{i=1}^{n-1} a_i^2 = A_2 \).

3. **Apply Cauchy-Schwarz inequality**: By the Cauchy-Schwarz inequality, we have:
   \[
   A_1^2 = \left( \sum_{i=1}^{n-1} a_i \cdot 1 \right)^2 \le (n-1) A_2
   \]

4. **Use given inequalities**: From the problem, we have:
   \[
   a_1 + a_2 + \cdots + a_n \ge n^2
   \]
   \[
   a_1^2 + a_2^2 + \cdots + a_n^2 \le n^3 + 1
   \]

5. **Express \( A_1 \) and \( A_2 \) in terms of \( a_n \)**: We can write:
   \[
   A_1 = \sum_{i=1}^{n-1} a_i = S - a_n \quad \text{where} \quad S = \sum_{i=1}^n a_i
   \]
   \[
   A_2 = \sum_{i=1}^{n-1} a_i^2 = T - a_n^2 \quad \text{where} \quad T = \sum_{i=1}^n a_i^2
   \]

6. **Substitute the inequalities**: Using the given inequalities, we get:
   \[
   S \ge n^2
   \]
   \[
   T \le n^3 + 1
   \]

7. **Apply Cauchy-Schwarz to \( A_1 \) and \( A_2 \)**:
   \[
   A_1^2 \le (n-1) A_2
   \]
   Substituting \( A_1 \) and \( A_2 \):
   \[
   (S - a_n)^2 \le (n-1)(T - a_n^2)
   \]

8. **Expand and simplify**:
   \[
   (S - a_n)^2 \le (n-1)(T - a_n^2)
   \]
   \[
   S^2 - 2Sa_n + a_n^2 \le (n-1)T - (n-1)a_n^2
   \]
   \[
   S^2 - 2Sa_n + a_n^2 \le (n-1)T - (n-1)a_n^2
   \]
   \[
   S^2 - 2Sa_n + n a_n^2 \le (n-1)T
   \]

9. **Substitute \( S \ge n^2 \) and \( T \le n^3 + 1 \)**:
   \[
   n^4 - 2n^2a_n + n a_n^2 \le (n-1)(n^3 + 1)
   \]
   \[
   n^4 - 2n^2a_n + n a_n^2 \le n^4 - n^3 + n^3 - 1
   \]
   \[
   n^4 - 2n^2a_n + n a_n^2 \le n^4 - 1
   \]

10. **Simplify further**:
    \[
    -2n^2a_n + n a_n^2 \le -1
    \]
    \[
    n a_n^2 - 2n^2a_n \le -1
    \]
    \[
    a_n^2 - 2n a_n \le -\frac{1}{n}
    \]

11. **Solve for \( a_n \)**: Let \( a_n = b + n \) where \( b \ge 0 \):
    \[
    (b+n)^2 - 2n(b+n) \le -\frac{1}{n}
    \]
    \[
    b^2 + 2bn + n^2 - 2nb - 2n^2 \le -\frac{1}{n}
    \]
    \[
    b^2 - n^2 \le -\frac{1}{n}
    \]

12. **Simplify and conclude**:
    \[
    b^2 \le n^2 - \frac{1}{n}
    \]
    Since \( b \ge 0 \), the only integer solution is \( b = 0 \). Thus, \( a_n = n \).

13. **Verify the solution**: Since \( a_n = n \) and \( a_1, a_2, \ldots, a_{n-1} \le a_n \), all \( a_i \) must be \( n \) to satisfy the sum condition \( \sum_{i=1}^n a_i \ge n^2 \).

14. **Conclusion**: The only solution is \( a_i = n \) for all \( i \).

The final answer is \( \boxed{ (a_1, a_2, \ldots, a_n) = (n, n, \ldots, n) } \).