If $a$ and $b$ are positive integers such that $a \cdot b = 2400,$ find the least possible value of $a + b.$
1. Given that \(a\) and \(b\) are positive integers such that \(a \cdot b = 2400\), we need to find the least possible value of \(a + b\).
2. To minimize \(a + b\), we need to find pairs \((a, b)\) such that \(a \cdot b = 2400\) and \(a\) and \(b\) are as close as possible.
3. Start by finding the prime factorization of 2400:
   \[
   2400 = 2^5 \times 3 \times 5^2
   \]
4. Next, we need to find pairs of factors of 2400 that are close to each other. We can list the factor pairs:
   \[
   \begin{aligned}
   &1 \times 2400, \\
   &2 \times 1200, \\
   &3 \times 800, \\
   &4 \times 600, \\
   &5 \times 480, \\
   &6 \times 400, \\
   &8 \times 300, \\
   &10 \times 240, \\
   &12 \times 200, \\
   &15 \times 160, \\
   &16 \times 150, \\
   &20 \times 120, \\
   &24 \times 100, \\
   &25 \times 96, \\
   &30 \times 80, \\
   &32 \times 75, \\
   &40 \times 60, \\
   &48 \times 50.
   \end{aligned}
   \]
5. Among these pairs, the pair \((48, 50)\) is the closest, and their sum is:
   \[
   48 + 50 = 98
   \]

The final answer is \(\boxed{98}\)