There's a real number written on every field of a $n \times n$ chess board. The sum of all numbers of a "cross" (union of a line and a column) is $\geq a$. What's the smallest possible sum of all numbers on the board¿
1. Let \( S \) be the sum of all numbers on the board.
2. Let \( S_{i,j} \) be the sum of the numbers in the union of row \( i \) and column \( j \).
3. Given that \( S_{i,j} \geq a \) for all \( 1 \leq i, j \leq n \).

We need to find the smallest possible sum \( S \) of all numbers on the board.

4. Consider the sum of all \( S_{i,j} \):
   \[
   \sum_{1 \leq i, j \leq n} S_{i,j}
   \]
   Each number on the board is counted once in its row sum and once in its column sum, except for the number at the intersection of the row and column, which is counted twice. Therefore, we have:
   \[
   \sum_{1 \leq i, j \leq n} S_{i,j} = (2n-1)S
   \]
5. Since \( S_{i,j} \geq a \) for all \( i, j \), we have:
   \[
   \sum_{1 \leq i, j \leq n} S_{i,j} \geq n^2 a
   \]
6. Combining the two equations, we get:
   \[
   (2n-1)S \geq n^2 a
   \]
7. Solving for \( S \), we find:
   \[
   S \geq \frac{n^2 a}{2n-1}
   \]

To check if this bound is achievable, consider filling the board with the value \( \frac{a}{2n-1} \) in each cell.

8. If each cell contains \( \frac{a}{2n-1} \), then the sum of any row or column is:
   \[
   n \cdot \frac{a}{2n-1}
   \]
9. The sum of the union of any row and column (a "cross") is:
   \[
   n \cdot \frac{a}{2n-1} + n \cdot \frac{a}{2n-1} - \frac{a}{2n-1} = \frac{2na - a}{2n-1} = a
   \]
   which satisfies the condition \( S_{i,j} \geq a \).

10. Therefore, the sum \( S \) of all numbers on the board is:
    \[
    n^2 \cdot \frac{a}{2n-1} = \frac{n^2 a}{2n-1}
    \]

Thus, the smallest possible sum of all numbers on the board is \( \frac{n^2 a}{2n-1} \).

The final answer is \( \boxed{\frac{n^2 a}{2n-1}} \).