What is the largest positive integer $n < 1000$ for which there is a positive integer $m$ satisfying \[\text{lcm}(m,n) = 3m \times \gcd(m,n)?\]

1. **Expressing \( m \) and \( n \) in terms of their greatest common divisor:**
   Let \( m = m'k \) and \( n = n'k \) where \( \gcd(m', n') = 1 \). This means \( k = \gcd(m, n) \).

2. **Using the properties of LCM and GCD:**
   We know that:
   \[
   \text{lcm}(m, n) = \frac{m \cdot n}{\gcd(m, n)}
   \]
   Substituting \( m = m'k \) and \( n = n'k \), we get:
   \[
   \text{lcm}(m, n) = \frac{(m'k) \cdot (n'k)}{k} = m'n'k
   \]

3. **Given condition:**
   According to the problem, we have:
   \[
   \text{lcm}(m, n) = 3m \times \gcd(m, n)
   \]
   Substituting the expressions for LCM and GCD, we get:
   \[
   m'n'k = 3(m'k) \times k
   \]
   Simplifying the right-hand side:
   \[
   m'n'k = 3m'k^2
   \]

4. **Simplifying the equation:**
   Dividing both sides by \( m'k \):
   \[
   n' = 3k
   \]

5. **Expressing \( n \) in terms of \( k \):**
   Since \( n = n'k \), substituting \( n' = 3k \):
   \[
   n = (3k)k = 3k^2
   \]

6. **Finding the largest \( n < 1000 \):**
   We need to find the largest integer \( k \) such that \( 3k^2 < 1000 \):
   \[
   k^2 < \frac{1000}{3} \approx 333.33
   \]
   Taking the integer part of the square root:
   \[
   k \leq \lfloor \sqrt{333.33} \rfloor = \lfloor 18.26 \rfloor = 18
   \]

7. **Calculating \( n \) for \( k = 18 \):**
   \[
   n = 3 \times 18^2 = 3 \times 324 = 972
   \]

Conclusion:
\[
\boxed{972}
\]