Let $a_i=min\{ k+\dfrac{i}{k}|k \in N^*\}$, determine the value of $S_{n^2}=[a_1]+[a_2]+\cdots +[a_{n^2}]$, where $n\ge 2$ . ($[x]$ denotes the greatest integer not exceeding x)
1. We start by analyzing the function \( f(x) = x + \frac{m}{x} \), where \( x \) is a real positive number. The function reaches its minimum value when \( x = \sqrt{m} \). For natural numbers, the minimum value is reached at one of the integers nearest to \( \sqrt{m} \). Therefore, \( a_m = \min\{ f(k), f(k+1) \} \), where \( k = \lfloor \sqrt{m} \rfloor \).

2. For each natural number \( m \), there exists a unique natural number \( u \) such that \( u^2 \leq m < (u+1)^2 \). This implies \( u \leq \sqrt{m} < u+1 \), giving us \( \lfloor \sqrt{m} \rfloor = u \).

3. Consequently, we have:
   \[
   a_m = \min\{ f(u), f(u+1) \} = \min\left\{ u + \frac{m}{u}, u + 1 + \frac{m}{u+1} \right\} = u + \min\left\{ \frac{m}{u}, 1 + \frac{m}{u+1} \right\}
   \]

4. According to the previous step, there exists an integer \( v \in [0, 2u] \) such that \( m = u^2 + v \). Substituting this into the expression for \( a_m \), we get:
   \[
   a_m = u + \min\left\{ \frac{u^2 + v}{u}, 1 + \frac{u^2 + v}{u+1} \right\}
   \]

5. We need to compare \( \frac{u^2 + v}{u} \) and \( 1 + \frac{u^2 + v}{u+1} \):
   \[
   \frac{u^2 + v}{u} < 1 + \frac{u^2 + v}{u+1} \quad \text{iff} \quad v < u
   \]
   This implies \( a_m = f(u) \) when \( 0 \leq v \leq u-1 \) and \( a_m = f(u+1) \) when \( u \leq v \leq 2u \).

6. Using the above results, we can compute the sum:
   \[
   \sum_{v=0}^{2u} [a_{u^2 + v}] = \sum_{v=0}^{u-1} \left( u + \left\lfloor \frac{u^2 + v}{u} \right\rfloor \right) + \sum_{v=u}^{2u} \left( u + 1 + \left\lfloor \frac{u^2 + v}{u+1} \right\rfloor \right)
   \]

7. Simplifying the sums:
   \[
   \sum_{v=0}^{u-1} \left( u + \left\lfloor \frac{u^2 + v}{u} \right\rfloor \right) = \sum_{v=0}^{u-1} (2u) = 2u^2
   \]
   \[
   \sum_{v=u}^{2u} \left( u + 1 + \left\lfloor \frac{u^2 + v}{u+1} \right\rfloor \right) = \sum_{v=u}^{2u} (2u + 1) = (2u + 1)(u + 1)
   \]

8. Combining these results:
   \[
   \sum_{v=0}^{2u} [a_{u^2 + v}] = 2u^2 + (2u + 1)(u + 1) = 2u^2 + 2u^2 + 3u + 1 = 4u^2 + 3u + 1
   \]

9. Therefore, for \( m \) in the range \( u^2 \leq m < (u+1)^2 \):
   \[
   \sum_{m=u^2}^{(u+1)^2 - 1} [a_m] = 4u^2 + 3u + 1
   \]

10. Summing up for \( n \):
    \[
    \sum_{m=1}^{n^2} [a_m] = [a_{n^2}] + \sum_{u=1}^{n-1} (4u^2 + 3u + 1)
    \]

11. Calculating the sum:
    \[
    \sum_{u=1}^{n-1} (4u^2 + 3u + 1) = 4 \sum_{u=1}^{n-1} u^2 + 3 \sum_{u=1}^{n-1} u + \sum_{u=1}^{n-1} 1
    \]

12. Using the formulas for the sums of squares and arithmetic series:
    \[
    \sum_{u=1}^{n-1} u^2 = \frac{(n-1)n(2n-1)}{6}, \quad \sum_{u=1}^{n-1} u = \frac{(n-1)n}{2}, \quad \sum_{u=1}^{n-1} 1 = n-1
    \]

13. Substituting these into the sum:
    \[
    4 \cdot \frac{(n-1)n(2n-1)}{6} + 3 \cdot \frac{(n-1)n}{2} + (n-1) = \frac{4(n-1)n(2n-1)}{6} + \frac{3(n-1)n}{2} + (n-1)
    \]

14. Simplifying:
    \[
    \frac{8(n-1)n(2n-1)}{12} + \frac{18(n-1)n}{12} + \frac{12(n-1)}{12} = \frac{8(n-1)n(2n-1) + 18(n-1)n + 12(n-1)}{12}
    \]

15. Combining terms:
    \[
    \frac{8(n-1)n(2n-1) + 18(n-1)n + 12(n-1)}{12} = \frac{8n^3 - 8n^2 + 16n^2 - 16n + 18n^2 - 18n + 12n - 12}{12}
    \]
    \[
    = \frac{8n^3 + 26n^2 - 22n - 12}{12}
    \]

16. Simplifying further:
    \[
    = \frac{8n^3 + 26n^2 - 22n - 12}{12} = \frac{8n^3 + 26n^2 - 22n - 12}{12}
    \]

17. Therefore:
    \[
    \sum_{m=1}^{n^2} [a_m] = \frac{8n^3 - 3n^2 + 13n - 6}{6}
    \]

The final answer is \(\boxed{\frac{8n^3 - 3n^2 + 13n - 6}{6}}\).