Calculate 

\[T = \sum \frac{1}{n_1! \cdot n_2! \cdot \cdots n_{1994}! \cdot (n_2 + 2 \cdot n_3 + 3 \cdot n_4 + \ldots + 1993 \cdot n_{1994})!}\]

where the sum is taken over all 1994-tuples of the numbers $n_1, n_2, \ldots, n_{1994} \in \mathbb{N} \cup \{0\}$ satisfying $n_1 + 2 \cdot n_2 + 3 \cdot n_3 + \ldots + 1994 \cdot n_{1994} = 1994.$
1. We start with the given sum:
   \[
   T = \sum \frac{1}{n_1! \cdot n_2! \cdots n_{1994}! \cdot (n_2 + 2 \cdot n_3 + 3 \cdot n_4 + \ldots + 1993 \cdot n_{1994})!}
   \]
   where the sum is taken over all 1994-tuples of the numbers \( n_1, n_2, \ldots, n_{1994} \in \mathbb{N} \cup \{0\} \) satisfying \( n_1 + 2 \cdot n_2 + 3 \cdot n_3 + \ldots + 1994 \cdot n_{1994} = 1994 \).

2. We can rewrite the constraint as:
   \[
   \sum_{i=1}^{1994} i \cdot n_i = 1994
   \]

3. Consider the generating function approach. The generating function for the sequence \( \{a_n\} \) where \( a_n \) is the number of ways to write \( n \) as a sum of the form \( \sum_{i=1}^{1994} i \cdot n_i \) is given by:
   \[
   \prod_{i=1}^{1994} \frac{1}{1-x^i}
   \]

4. We need to find the coefficient of \( x^{1994} \) in the expansion of:
   \[
   \prod_{i=1}^{1994} \frac{1}{1-x^i}
   \]

5. This coefficient is known to be the number of partitions of 1994 into parts where each part is at most 1994. This is a well-known combinatorial problem, and the number of such partitions is given by the partition function \( p(1994) \).

6. The sum \( T \) can be interpreted as the sum of the reciprocals of the factorials of the parts in each partition of 1994. This is equivalent to the coefficient of \( x^{1994} \) in the expansion of:
   \[
   \prod_{i=1}^{1994} \sum_{n_i=0}^{\infty} \frac{x^{i \cdot n_i}}{n_i!}
   \]

7. This simplifies to:
   \[
   \prod_{i=1}^{1994} e^{x^i} = e^{\sum_{i=1}^{1994} x^i} = e^{\frac{x(1-x^{1994})}{1-x}}
   \]

8. We need the coefficient of \( x^{1994} \) in the expansion of \( e^{\frac{x(1-x^{1994})}{1-x}} \). This is equivalent to finding the coefficient of \( x^{1994} \) in:
   \[
   e^{\frac{x}{1-x}}
   \]

9. The series expansion of \( e^{\frac{x}{1-x}} \) is:
   \[
   e^{\frac{x}{1-x}} = \sum_{n=0}^{\infty} \frac{1}{n!} \left( \frac{x}{1-x} \right)^n
   \]

10. The coefficient of \( x^{1994} \) in this expansion is given by:
    \[
    \frac{1}{1994!}
    \]

Conclusion:
\[
T = \frac{1}{1994!}
\]

The final answer is \(\boxed{\frac{1}{1994!}}\)