Find all integers $n\ge 3$ for which the following statement is true:
Any arithmetic progression $a_1,\ldots ,a_n$ with $n$ terms for which $a_1+2a_2+\ldots+na_n$ is rational contains at least one rational term.
1. **Define the arithmetic progression and the given condition:**
   Let the arithmetic progression be \(a_1, a_2, \ldots, a_n\) with common difference \(d\). The terms of the progression can be written as:
   \[
   a_1, a_1 + d, a_1 + 2d, \ldots, a_1 + (n-1)d
   \]
   The given condition is that the sum \(a_1 + 2a_2 + 3a_3 + \ldots + na_n\) is rational.

2. **Express the sum \(a_1 + 2a_2 + 3a_3 + \ldots + na_n\):**
   \[
   a_1 + 2(a_1 + d) + 3(a_1 + 2d) + \ldots + n(a_1 + (n-1)d)
   \]
   Simplify the expression:
   \[
   a_1 + 2a_1 + 2d + 3a_1 + 6d + \ldots + na_1 + n(n-1)d
   \]
   Combine like terms:
   \[
   (1 + 2 + 3 + \ldots + n)a_1 + (2 \cdot 1 + 3 \cdot 2 + \ldots + n \cdot (n-1))d
   \]

3. **Simplify the sums:**
   The sum of the first \(n\) natural numbers is:
   \[
   1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}
   \]
   The sum of the products \(2 \cdot 1 + 3 \cdot 2 + \ldots + n \cdot (n-1)\) can be written as:
   \[
   \sum_{k=1}^{n} k(k-1) = \sum_{k=1}^{n} (k^2 - k) = \left(\sum_{k=1}^{n} k^2\right) - \left(\sum_{k=1}^{n} k\right)
   \]
   Using the formulas for the sum of squares and the sum of the first \(n\) natural numbers:
   \[
   \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   \[
   \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
   \]
   Therefore:
   \[
   \sum_{k=1}^{n} k(k-1) = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)}{6} (2n+1 - 3) = \frac{n(n+1)(2n-2)}{6} = \frac{n(n+1)(n-1)}{3}
   \]

4. **Combine the results:**
   \[
   a_1 + 2a_2 + 3a_3 + \ldots + na_n = \frac{n(n+1)}{2}a_1 + \frac{n(n+1)(n-1)}{3}d
   \]
   Factor out \(\frac{n(n+1)}{6}\):
   \[
   \frac{n(n+1)}{6} (3a_1 + 2(n-1)d)
   \]
   For this expression to be rational, \(3a_1 + 2(n-1)d\) must be rational.

5. **Analyze the condition \(3a_1 + 2(n-1)d\) being rational:**
   - If \(3 \mid (n-1)\), then \(n-1\) is a multiple of 3, say \(n-1 = 3k\). Thus, \(n = 3k + 1\).
   - In this case, \(3a_1 + 2 \cdot 3k \cdot d = 3a_1 + 6kd\) is rational if \(a_1\) and \(d\) are rational.

6. **Show that for other \(n\), the condition fails:**
   - If \(n \not\equiv 1 \pmod{3}\), then \(n-1\) is not a multiple of 3.
   - Let \(d = \frac{-3a_1}{2(n-1)}\) and \(a_1\) be an irrational number. Then all terms of the progression are non-zero rational multiples of \(a_1\), making them irrational.

Conclusion:
The condition holds if and only if \(n \equiv 1 \pmod{3}\).

The final answer is \( \boxed{ n \equiv 1 \pmod{3} } \).