For any given integers $m,n$ such that $2\leq m<n$ and $(m,n)=1$. Determine the smallest positive integer $k$ satisfying the following condition: for any $m$-element subset $I$ of $\{1,2,\cdots,n\}$ if $\sum_{i\in I}i> k$, then there exists a sequence of $n$ real numbers $a_1\leq a_2 \leq \cdots \leq a_n$ such that

$$\frac1m\sum_{i\in I} a_i>\frac1n\sum_{i=1}^na_i$$
1. **Assumption and Simplification**:
   - We start by noting that the sequence \(a_1, a_2, \ldots, a_n\) satisfies the given conditions if and only if the sequence \((a_1 + w, a_2 + w, \ldots, a_n + w)\) also satisfies the conditions for any real number \(w\). Therefore, we can assume without loss of generality that \(a_1 = 0\).

2. **Subset and Condition**:
   - Consider any \(m\)-element subset \(I = \{i_1, i_2, \ldots, i_m\}\) of \(\{1, 2, \ldots, n\}\) with \(i_1 < i_2 < \cdots < i_m\). We need to ensure that for any such subset, if \(\sum_{i \in I} i > k\), then \(\frac{1}{m} \sum_{i \in I} a_i > \frac{1}{n} \sum_{i=1}^n a_i\).

3. **Rewriting the Condition**:
   - Given \(a_1 \leq a_2 \leq \cdots \leq a_n\), we can rewrite the sequence as \((0, 0, \ldots, 0, a_{i_1}, a_{i_1}, \ldots, a_{i_1}, a_{i_2}, a_{i_2}, \ldots, a_{i_2}, \ldots, a_{i_m}, a_{i_m}, \ldots, a_{i_m})\).

4. **Inequality Transformation**:
   - We need to show that:
     \[
     \frac{a_{i_1} + a_{i_2} + \cdots + a_{i_m}}{m} \geq \frac{\left( \sum_{j=1}^{m-1} a_{i_j} (i_{j+1} - i_j) \right) + a_{i_m} (n - i_m + 1)}{n}
     \]
     Simplifying, we get:
     \[
     \left( \sum_{j=1}^{m-1} a_{i_j} (n - m (i_{j+1} - i_j)) \right) + a_{i_m} (n - m (n - i_m + 1)) \geq 0
     \]

5. **Introducing Differences**:
   - Let \(t_{i_k} = a_{i_k} - a_{i_{k-1}} > 0\) for \(k = 2, 3, \ldots, m\). The inequality reduces to:
     \[
     a_{i_1} (mn - m (n - i_1 + 1)) + \sum_{j=2}^{m} t_{i_j} ((m - (j-1)) n - m (n - i_j + 1)) \geq 0
     \]
     where \(t_{i_2}, t_{i_3}, \ldots, t_{i_m} > 0\) and \(a_{i_1} \geq 0\).

6. **Non-existence Condition**:
   - There does not exist such \(a_{i_1}, t_{i_1}, t_{i_2}, \ldots, t_{i_m}\) if and only if:
     \[
     mn - m (n - i_1 + 1) < 0 \quad \text{and} \quad (m - (j-1)) n - m (n - i_j + 1) < 0 \quad \text{for all } j = 2, 3, \ldots, m
     \]

7. **Choosing Indices**:
   - This means that if \(i_1 = 1 = \left\lfloor \frac{n(1-1)}{m} + 1 \right\rfloor\) and \(i_j = \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor\) for all \(j = 2, 3, \ldots, m\), then there does not exist \(a_1, a_2, \ldots, a_n\).

8. **Finding \(k\)**:
   - Therefore, \(k \geq 1 + \sum_{j=2}^{m} \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor\).
   - For these values of \(k\), we get that \(\sum_{j=1}^{m} \left( i_j - \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor \right) > 0\).
   - This implies there exists \(h \in \{1, 2, \ldots, m\}\) such that \(i_h > \left\lfloor \frac{n(h-1)}{m} + 1 \right\rfloor\), then choosing \(t_{i_h} \rightarrow \infty\) makes the inequality hold.

9. **Smallest Value of \(k\)**:
   - Hence, the smallest value of \(k\) is:
     \[
     \sum_{j=1}^{m} \left( \left\lfloor \frac{n(j-1)}{m} + 1 \right\rfloor \right) = \sum_{j=1}^{m} \left( 1 + \frac{n(j-1)}{m} \right) - \left( \frac{0}{m} + \frac{1}{m} + \cdots + \frac{m-1}{m} \right) = \frac{(n-1)(m-1)}{2} + m
     \]

The final answer is \(\boxed{\frac{(n-1)(m-1)}{2} + m}\)