For each positive integer $ k$, find the smallest number $ n_{k}$ for which there exist real $ n_{k}\times n_{k}$ matrices $ A_{1}, A_{2}, \ldots, A_{k}$ such that all of the following conditions hold:

(1) $ A_{1}^{2}= A_{2}^{2}= \ldots = A_{k}^{2}= 0$,

(2) $ A_{i}A_{j}= A_{j}A_{i}$ for all $ 1 \le i, j \le k$, and 

(3) $ A_{1}A_{2}\ldots A_{k}\ne 0$.
To find the smallest number \( n_k \) for which there exist real \( n_k \times n_k \) matrices \( A_1, A_2, \ldots, A_k \) satisfying the given conditions, we need to establish both a lower bound and an upper bound for \( n_k \).

### Step 1: Establishing the Lower Bound \( n_k \geq 2^k \)

We will use induction to show that \( n_k \geq 2^k \).

**Base Case: \( k = 1 \)**

For \( k = 1 \), we need a matrix \( A_1 \) such that \( A_1^2 = 0 \) and \( A_1 \neq 0 \). The smallest such matrix is a \( 2 \times 2 \) nilpotent matrix, for example:
\[ A_1 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \]
Thus, \( n_1 = 2 \), and \( 2 = 2^1 \).

**Inductive Step:**

Assume that for some \( k \geq 1 \), \( n_k \geq 2^k \). We need to show that \( n_{k+1} \geq 2^{k+1} \).

Consider \( n_{k+1} \times n_{k+1} \) matrices \( A_1, A_2, \ldots, A_{k+1} \) satisfying the conditions. Let \( V \) be an \( L \)-vector space of dimension \( n_{k+1} \) on which the \( A_i \) act. The kernel of \( A_1 \), denoted \( \ker(A_1) \), has dimension at least \( \frac{n_{k+1}}{2} \) because \( A_1^2 = 0 \) implies that the range of \( A_1 \) is contained in \( \ker(A_1) \).

Since all \( A_i \) commute, \( \ker(A_1) \) is invariant under all \( A_i \). Let \( V' = V / \ker(A_1) \), which has dimension at most \( \frac{n_{k+1}}{2} \). The matrices \( A_2, \ldots, A_{k+1} \) act on \( V' \) and satisfy the same conditions. By the induction hypothesis, the dimension of \( V' \) must be at least \( 2^k \). Therefore:
\[ \frac{n_{k+1}}{2} \geq 2^k \]
\[ n_{k+1} \geq 2^{k+1} \]

Thus, by induction, \( n_k \geq 2^k \) for all \( k \).

### Step 2: Establishing the Upper Bound \( n_k \leq 2^k \)

We need to construct \( n_k \times n_k \) matrices \( A_1, A_2, \ldots, A_k \) that satisfy the conditions with \( n_k = 2^k \).

Consider the algebra \( \mathcal{A} = L[x_1, x_2, \ldots, x_k] / (x_1^2, x_2^2, \ldots, x_k^2) \). This algebra has dimension \( 2^k \) over \( L \), with a basis consisting of all products of distinct \( x_i \) (including the empty product, which is the multiplicative unit).

Define \( A_i \) to be the matrix representation of multiplication by \( x_i \) in \( \mathcal{A} \). These matrices satisfy:
1. \( A_i^2 = 0 \) because \( x_i^2 = 0 \) in \( \mathcal{A} \).
2. \( A_i A_j = A_j A_i \) because multiplication in \( \mathcal{A} \) is commutative.
3. \( A_1 A_2 \ldots A_k \neq 0 \) because the product \( x_1 x_2 \ldots x_k \) is a non-zero element in \( \mathcal{A} \).

Thus, \( n_k = 2^k \) is achievable.

### Conclusion

Combining the lower and upper bounds, we conclude that the smallest \( n_k \) is \( 2^k \).

The final answer is \( \boxed{ n_k = 2^k } \).