If $k, l, m$ are positive integers with $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}<1$, ﬁnd the maximum possible value of $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}$.
1. We start with the given inequality:
   \[
   \frac{1}{k} + \frac{1}{l} + \frac{1}{m} < 1
   \]
   where \( k, l, m \) are positive integers.

2. To find the maximum possible value of \(\frac{1}{k} + \frac{1}{l} + \frac{1}{m}\), we need to consider the constraints and properties of the harmonic series. 

3. Notice that if \( k, l, m \) are all greater than or equal to 3, then:
   \[
   \frac{1}{k} + \frac{1}{l} + \frac{1}{m} \leq \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1
   \]
   which does not satisfy the inequality. Therefore, at least one of \( k, l, m \) must be 2.

4. Without loss of generality, let \( k = 2 \). Then the inequality becomes:
   \[
   \frac{1}{2} + \frac{1}{l} + \frac{1}{m} < 1
   \]
   Simplifying, we get:
   \[
   \frac{1}{l} + \frac{1}{m} < \frac{1}{2}
   \]

5. To maximize \(\frac{1}{l} + \frac{1}{m}\), we need to find the smallest possible values of \( l \) and \( m \) that satisfy the inequality. 

6. Let's try \( l = 3 \). Then we have:
   \[
   \frac{1}{3} + \frac{1}{m} < \frac{1}{2}
   \]
   Simplifying, we get:
   \[
   \frac{1}{m} < \frac{1}{2} - \frac{1}{3} = \frac{1}{6}
   \]
   Therefore, \( m \) must be greater than 6. The smallest integer greater than 6 is 7. So, let \( m = 7 \).

7. Now, we have:
   \[
   \frac{1}{k} + \frac{1}{l} + \frac{1}{m} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7}
   \]

8. To find the sum:
   \[
   \frac{1}{2} + \frac{1}{3} + \frac{1}{7} = \frac{21}{42} + \frac{14}{42} + \frac{6}{42} = \frac{41}{42}
   \]

9. We need to verify if this is indeed the maximum value. If we try other combinations, such as \( l = 4 \) or \( l = 5 \), we find that the resulting sums are less than \(\frac{41}{42}\).

Therefore, the maximum possible value of \(\frac{1}{k} + \frac{1}{l} + \frac{1}{m}\) is indeed \(\frac{41}{42}\).

The final answer is \(\boxed{\frac{41}{42}}\).