At first suppose, that number $n = p_1^{m_1} \dots p_k^{m_k}$, where $p_i$, $i = 1, \overline{k}$ – are distinct prime numbers and $m_i$ – are natural numbers, $i = 1, \overline{k}$. Then let's choose all ordered pairs of numbers $(a, b)$, for which $[a, b] = n$. Here we do not have condition $a < b$, so pairs $(a, b)$ and $(b, a)$ for $a \neq b$ are considered different. If $a = p_1^{a_1} \dots p_k^{a_k}$ and $b = p_1^{b_1} \dots p_k^{b_k}$, so for satisfying condition $[a, b] = n$ it is necessary and sufficient that $\forall i = 1, \overline{k} \max\{a_i, b_i\} = m_i$. Then possible variants of pairs $(a_i, b_i)$:
$(0, m_i), (1, m_i), \dots, (m_i - 1, m_i), (m_i, m_i), (m_i, m_i - 1), (m_i, m_i - 2), \dots, (m_i, 0).$
In total there are $2m_i + 1$ variants. So, in total there will be exactly $N = (2m_1 + 1) \dots (2m_k + 1)$ ordered pairs of numbers. And there are exactly $\frac{1}{2}(N-1)$ pairs, for which $a < b$. Really, for all computed pairs we need to delete pair $(n, n)$, and for every two pairs $(a, b)$ and $(b, a)$ we need to leave one. So, we need to find the biggest three-digit natural number, for which $\frac{1}{2}(N-1) = 16$.

Then $N = 33$, so there are two possible variants: $N = 2 \cdot 16 + 1$ or $N = (2 \cdot 5 + 1)(2 \cdot 1 + 1)$. In first case, for some prime $p$ condition is true $n = p^{16} \ge 2^{16} > 999$. In second case, we have, that $n = p^5 q$, where $p, q$ – are distinct prime numbers. As $5^5 > 999$, then $p = 2$ or $p = 3$.
For $p=3$ we have, that $3^5 = 243$, so the number can only be $3^5 \cdot 2 = 486$, as $3^5 \cdot 5 > 999$.
For $p=2$ we have, that $2^5 = 32$, so the number can be maximum three-digit number of form: $2^5 \cdot p$, where $p > 2$ – is prime. As $1000: 2^5 = 31.25$, it is not hard to see, that the number is $2^5 \cdot 31 = 992$.