Determine all positive integers $k$, $\ell$, $m$ and $n$, such that $$\frac{1}{k!}+\frac{1}{\ell!}+\frac{1}{m!} =\frac{1}{n!} $$
To determine all positive integers \( k, \ell, m, \) and \( n \) such that 
\[ \frac{1}{k!} + \frac{1}{\ell!} + \frac{1}{m!} = \frac{1}{n!}, \]
we start by noting that \( n \) must be less than \( k, \ell, \) and \( m \) because the sum of three positive terms on the left-hand side must be greater than any single term on the right-hand side.

1. **Case \( k = \ell = m \):**
   \[ \frac{3}{k!} = \frac{1}{n!}. \]
   This implies:
   \[ 3 = \frac{k!}{n!}. \]
   Since \( k! \) must be divisible by \( n! \), and \( 3 \) is a small number, we test small values of \( k \) and \( n \):
   - If \( k = 3 \), then \( k! = 6 \). We need \( \frac{6}{n!} = 3 \), which gives \( n! = 2 \). Thus, \( n = 2 \).
   - Therefore, one solution is \( (k, \ell, m, n) = (3, 3, 3, 2) \).

2. **Case \( k \neq \ell \neq m \):**
   We need to consider different values for \( k, \ell, \) and \( m \). We assume \( k \leq \ell \leq m \).

   - **Subcase \( k - n = 1 \):**
     \[ \frac{3}{k!} = \frac{1}{(k-1)!}. \]
     This implies:
     \[ 3 = k, \]
     which we have already considered.

   - **Subcase \( k - n = 2 \):**
     We have three subcases to consider:
     - **Subcase 1:**
       \[ \frac{1}{k!} + \frac{1}{(k-1)!} + \frac{1}{k!} = \frac{1}{(k-2)!}. \]
       Simplifying, we get:
       \[ \frac{2}{k!} + \frac{1}{(k-1)!} = \frac{1}{(k-2)!}. \]
       \[ \frac{2}{k!} + \frac{k}{k!} = \frac{1}{(k-2)!}. \]
       \[ \frac{2 + k}{k!} = \frac{1}{(k-2)!}. \]
       \[ 2 + k = k(k-1). \]
       Solving this, we get:
       \[ k^2 - 2k - 2 = 0. \]
       This quadratic equation does not have integer solutions.

     - **Subcase 2:**
       \[ \frac{1}{k!} + \frac{1}{(k-1)!} + \frac{1}{(k-1)!} = \frac{1}{(k-2)!}. \]
       Simplifying, we get:
       \[ \frac{1}{k!} + \frac{2}{(k-1)!} = \frac{1}{(k-2)!}. \]
       \[ \frac{1}{k!} + \frac{2k}{k!} = \frac{1}{(k-2)!}. \]
       \[ \frac{1 + 2k}{k!} = \frac{1}{(k-2)!}. \]
       \[ 1 + 2k = k(k-1). \]
       Solving this, we get:
       \[ k^2 - 3k - 1 = 0. \]
       This quadratic equation does not have integer solutions.

     - **Subcase 3:**
       \[ \frac{3}{k!} = \frac{1}{(k-2)!}. \]
       This implies:
       \[ 3 = k(k-1). \]
       Solving this, we get:
       \[ k^2 - k - 3 = 0. \]
       This quadratic equation does not have integer solutions.

3. **Case \( k - n \geq 3 \):**
   Assume \( m \leq \ell \leq k \). Then we have:
   \[ 1 + \frac{k!}{\ell!} + \frac{k!}{m!} = \frac{k!}{n!}. \]
   Since the right-hand side is even, we have \( \ell = k-1 \). But then \( k \) must divide the right-hand side, so \( m = k \), which contradicts \( m \leq \ell \).

Thus, the only solution is:
\[ \boxed{(k, \ell, m, n) = (3, 3, 3, 2)}. \]