Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer.
1. Let \(a, b, c\) be three distinct positive integers such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer. Without loss of generality (WLOG), assume \(a > b > c\).

2. We claim that \(a = 6\), \(b = 3\), and \(c = 2\) is a solution. First, we verify that this set satisfies the given conditions:
   - The sum of the reciprocals of \(a\) and \(b\) is \(\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\), which is an integral multiple of \(\frac{1}{2}\).
   - The sum of the reciprocals of \(a\) and \(c\) is \(\frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}\), which is an integral multiple of \(\frac{1}{3}\).
   - The sum of the reciprocals of \(b\) and \(c\) is \(\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}\), which is an integral multiple of \(\frac{1}{6}\).

3. To prove our claim, we need to show that if \((a, b, c) \neq (6, 3, 2)\), then \(a + b + c\) is at least 12. Note that \(6 + 3 + 2 = 11\).

4. Since \(a > b > c\), we have \(\frac{1}{a} < \frac{1}{b} < \frac{1}{c}\). This implies \(\frac{1}{a} + \frac{1}{b} < \frac{2}{c}\).

5. Given that \(\frac{1}{a} + \frac{1}{b}\) is an integral multiple of \(\frac{1}{c}\), we can write:
   \[
   \frac{1}{a} + \frac{1}{b} = k \cdot \frac{1}{c}
   \]
   for some integer \(k\). Since \(\frac{1}{a} + \frac{1}{b} < \frac{2}{c}\), it follows that \(k = 1\), so:
   \[
   \frac{1}{a} + \frac{1}{b} = \frac{1}{c}
   \]

6. Consider the case \(c = 2\):
   - If \(b = 3\), then \(\frac{1}{a} + \frac{1}{3} = \frac{1}{2}\). Solving for \(a\), we get:
     \[
     \frac{1}{a} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} \implies a = 6
     \]
     Thus, \((a, b, c) = (6, 3, 2)\), which is our original solution.
   - If \(b = 4\), then \(\frac{1}{a} + \frac{1}{4} = \frac{1}{2}\). Solving for \(a\), we get:
     \[
     \frac{1}{a} = \frac{1}{2} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4} \implies a = 4
     \]
     This leads to \(a = b = 4\), which contradicts \(a > b\). Therefore, \(b \geq 5\), implying \(a \geq 6\). Thus, \(a + b + c \geq 6 + 5 + 2 = 13\).

7. Consider the case \(c \geq 3\):
   - Then \(b \geq 4\) and \(a \geq 5\). Therefore, \(a + b + c \geq 5 + 4 + 3 = 12\).

8. In both cases, if \((a, b, c) \neq (6, 3, 2)\), then \(a + b + c \geq 12\).

Thus, the smallest sum of three distinct positive integers satisfying the given conditions is \(6 + 3 + 2 = 11\).

The final answer is \(\boxed{11}\).