For each positive integer $n \ge 3$, define $A_n$ and $B_n$ as
\[A_n = \sqrt{n^2 + 1} + \sqrt{n^2 + 3} + \cdots + \sqrt{n^2+2n-1}\]
\[B_n = \sqrt{n^2 + 2} + \sqrt{n^2 + 4} + \cdots + \sqrt{n^2 + 2n}.\]
Determine all positive integers $n\ge 3$ for which $\lfloor A_n \rfloor = \lfloor B_n \rfloor$.
Note. For any real number $x$, $\lfloor x\rfloor$ denotes the largest integer $N\le x$.

[i]Anant Mudgal and Navilarekallu Tejaswi[/i]
1. **Define \(A_n\) and \(B_n\):**
   \[
   A_n = \sqrt{n^2 + 1} + \sqrt{n^2 + 3} + \cdots + \sqrt{n^2 + 2n - 1}
   \]
   \[
   B_n = \sqrt{n^2 + 2} + \sqrt{n^2 + 4} + \cdots + \sqrt{n^2 + 2n}
   \]

2. **Observation:**
   For \(n^2 < n^2 + k < (n+1)^2\), we have:
   \[
   \lfloor \sqrt{n^2 + k} \rfloor = n
   \]

3. **Claim:**
   We claim that the following holds:
   \[
   \frac{k-1}{2n} < \left\{\sqrt{n^2 + k}\right\} < \frac{k}{2n} \quad \forall \, n \geq 3, \quad k \in \{1, 2, \ldots, 2n\}
   \]

4. **Proof of the Claim:**
   - **Lower Bound:**
     \[
     \{\sqrt{n^2 + k}\} > \frac{k-1}{2n}
     \]
     This is equivalent to:
     \[
     \sqrt{n^2 + k} > n + \frac{k-1}{2n}
     \]
     Squaring both sides:
     \[
     n^2 + k > n^2 + \frac{(k-1)^2}{4n^2} + k - 1
     \]
     Simplifying:
     \[
     \frac{(2n)^2}{(k-1)^2} > 1
     \]
     which is obvious.

   - **Upper Bound:**
     \[
     \sqrt{n^2 + k} < n + \frac{k}{2n}
     \]
     Squaring both sides:
     \[
     n^2 + k < n^2 + \frac{k^2}{4n^2} + k
     \]
     which is again obvious.

5. **Calculating \(\lfloor A_n \rfloor\):**
   \[
   \left\lfloor A_n \right\rfloor = \left\lfloor \sqrt{n^2 + 1} + \sqrt{n^2 + 3} + \cdots + \sqrt{n^2 + 2n-1} \right\rfloor
   \]
   \[
   = \left\lfloor \left\lfloor \sqrt{n^2 + 1} \right\rfloor + \left\lfloor \sqrt{n^2 + 3} \right\rfloor + \cdots + \left\lfloor \sqrt{n^2 + 2n-1} \right\rfloor + \left\{ \sqrt{n^2 + 1} \right\} + \cdots + \left\{ \sqrt{n^2 + 2n-1} \right\} \right\rfloor
   \]
   \[
   = \lfloor \underbrace{n + n + \cdots + n}_{n \text{ times}} + \{ \sqrt{n^2 + 1} \} + \cdots + \{ \sqrt{n^2 + 2n-1} \} \rfloor
   \]
   \[
   = n^2 + \left\lfloor \left\{ \sqrt{n^2 + 1} \right\} + \cdots + \left\{ \sqrt{n^2 + 2n-1} \right\} \right\rfloor
   \]
   \[
   = n^2 + \lfloor X \rfloor
   \]

6. **Calculating \(\lfloor B_n \rfloor\):**
   \[
   \left\lfloor B_n \right\rfloor = \left\lfloor \sqrt{n^2 + 2} + \sqrt{n^2 + 4} + \cdots + \sqrt{n^2 + 2n} \right\rfloor
   \]
   \[
   = \left\lfloor \left\lfloor \sqrt{n^2 + 2} \right\rfloor + \left\lfloor \sqrt{n^2 + 4} \right\rfloor + \cdots + \left\lfloor \sqrt{n^2 + 2n} \right\rfloor + \left\{ \sqrt{n^2 + 2} \right\} + \cdots + \left\{ \sqrt{n^2 + 2n} \right\} \right\rfloor
   \]
   \[
   = \lfloor \underbrace{n + n + \cdots + n}_{n \text{ times}} + \{ \sqrt{n^2 + 2} \} + \cdots + \{ \sqrt{n^2 + 2n} \} \rfloor
   \]
   \[
   = n^2 + \left\lfloor \left\{ \sqrt{n^2 + 2} \right\} + \cdots + \left\{ \sqrt{n^2 + 2n} \right\} \right\rfloor
   \]
   \[
   = n^2 + \lfloor Y \rfloor
   \]

7. **Equating \(\lfloor A_n \rfloor\) and \(\lfloor B_n \rfloor\):**
   \[
   \lfloor A_n \rfloor = \lfloor B_n \rfloor \implies n^2 + \lfloor X \rfloor = n^2 + \lfloor Y \rfloor \implies \lfloor X \rfloor = \lfloor Y \rfloor
   \]

8. **Bounds for \(X\) and \(Y\):**
   \[
   \frac{1}{2n} \sum_{k \text{ odd}} (k-1) < X < \frac{1}{2n} \sum_{k \text{ odd}} k
   \]
   \[
   \frac{1}{2n}(0 + 2 + \cdots + 2n-2) < X < \frac{1}{2n}(1 + 3 + \cdots + 2n-1)
   \]
   \[
   \frac{1}{n} \frac{(n-1) \cdot n}{2} < X < \frac{n^2}{2n}
   \]
   \[
   \frac{n-1}{2} < X < \frac{n}{2}
   \]

   Similarly for \(Y\):
   \[
   \frac{1}{2n} \sum_{k \text{ even}} (k-1) < Y < \frac{1}{2n} \sum_{k \text{ even}} k
   \]
   \[
   \frac{1}{2n} 2(1 + 3 + \cdots + 2n-1) < Y < \frac{1}{2n} 2(1 + \cdots + n)
   \]
   \[
   \frac{n}{2} < Y < \frac{n+1}{2}
   \]

9. **Case Analysis:**
   - **If \(n\) is even, say \(n = 2k\):**
     \[
     \frac{2k-1}{2} < X < \frac{2k}{2} \implies \lfloor X \rfloor = k-1
     \]
     \[
     \frac{2k}{2} < Y < \frac{2k+1}{2} \implies \lfloor Y \rfloor = k
     \]
     This is not possible since \(\lfloor X \rfloor \neq \lfloor Y \rfloor\).

   - **If \(n\) is odd, say \(n = 2k+1\):**
     \[
     \frac{2k}{2} < X < \frac{2k+1}{2} \implies \lfloor X \rfloor = k
     \]
     \[
     \frac{2k+1}{2} < Y < \frac{2k+2}{2} \implies \lfloor Y \rfloor = k
     \]
     This is possible since \(\lfloor X \rfloor = \lfloor Y \rfloor\).

10. **Conclusion:**
    The equality \(\lfloor A_n \rfloor = \lfloor B_n \rfloor\) holds for all odd numbers \(n \geq 3\).

The final answer is \( \boxed{ n } \) is odd and \(n \geq 3\).