I take the \(1729\)-digit number \(\overline{a_{1}a_{2} \ldots a_{1729}}\) and concantenate it with itself \(1729\) times to make a new \(1729^{2}\)-digit number

\[N = \underbrace{\overline{a_{1}a_{2} \ldots a_{1729}a_{1}a_{2} \ldots a_{1729} \ldots \ldots a_{1}a_{2} \ldots a_{1729}}}_{1729 \ \text{times}}\]

If the sequence \(\lbrace a_{i} \rbrace\) satisfies that \(\displaystyle \sum_{i=1}^{1729} a_{i}^{2} = 49000\) and there are exactly \(490\) solutions \(i\) to the congruence \(a_{i} \equiv 2 \pmod{3}\), find the number of \(N\) divisible by \(3\).

Note: \(0 \leq a_{i} \leq 9\).

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