Let $ \{m_1,m_2,\dots\}$ be a (finite or infinite) set of positive integers. Consider the system of congruences 

(1)   $ x\equiv 2m_i^2 \pmod{2m_i\minus{}1}$ ($ i\equal{}1,2,...$ ).

Give a necessary and sufficient condition for the system (1) to be solvable.
1. We start with the given system of congruences:
   \[
   x \equiv 2m_i^2 \pmod{2m_i - 1} \quad \text{for} \quad i = 1, 2, \ldots
   \]
   We need to find a necessary and sufficient condition for this system to be solvable.

2. First, observe that:
   \[
   2m_i^2 \equiv m_i \pmod{2m_i - 1}
   \]
   This is because:
   \[
   2m_i^2 = m_i + (2m_i - 1)k \quad \text{for some integer} \quad k
   \]
   Hence:
   \[
   x \equiv m_i \pmod{2m_i - 1}
   \]

3. Next, we need to ensure that \( x \) satisfies all the congruences simultaneously. This means:
   \[
   x \equiv m_i \pmod{2m_i - 1} \quad \text{for all} \quad i
   \]

4. To solve this, we consider the least common multiple (LCM) of the moduli \( 2m_i - 1 \). Let:
   \[
   k = \text{lcm}(2m_1 - 1, 2m_2 - 1, \ldots)
   \]

5. We need \( x \) to satisfy:
   \[
   x \equiv m_i \pmod{2m_i - 1} \quad \text{for all} \quad i
   \]
   This implies:
   \[
   2x \equiv 2m_i \pmod{2m_i - 1} \equiv 1 \pmod{2m_i - 1}
   \]

6. Let \( p_j^{e_j} \) be the largest power of \( p_j \) dividing any of the \( 2m_i - 1 \) across all possible \( i \). We then have:
   \[
   2x \equiv 1 \pmod{p_j^{e_j}} \quad \text{for all} \quad j
   \]

7. Since \( p_j^{e_j} \) are distinct prime powers, they are pairwise relatively prime. By the Chinese Remainder Theorem, there exists exactly one value of \( 2x \pmod{k} \) that satisfies the conditions. Therefore:
   \[
   2x \equiv 1 \pmod{k}
   \]

8. Solving for \( x \), we get:
   \[
   x \equiv \frac{k + 1}{2} \pmod{k}
   \]

9. This solution is only finite if \( k \) is finite, which means the sequence \( 2m_i - 1 \) must have a finite LCM.

Conclusion:
The necessary and sufficient condition for the system of congruences to be solvable is that the sequence \( 2m_i - 1 \) has a finite least common multiple (LCM).

The final answer is the sequence \( \boxed{ 2m_i - 1 } \) must have a finite least common multiple (LCM).