The irrational numbers $\alpha ,\beta ,\gamma ,\delta$ are such that $\forall$ $n\in \mathbb{N}$ :
$[n\alpha ].[n\beta ]=[n\gamma ].[n\delta ]$.
Is it true that the sets $\{ \alpha ,\beta \}$ and $\{ \gamma ,\delta \}$ are equal?

1. **Given Condition**: We start with the given condition for all \( n \in \mathbb{N} \):
   \[
   [n\alpha] \cdot [n\beta] = [n\gamma] \cdot [n\delta]
   \]
   Here, \([x]\) denotes the floor function, which gives the greatest integer less than or equal to \( x \).

2. **Asymptotic Behavior**: As \( n \to \infty \), the floor functions \([n\alpha]\) and \([n\beta]\) can be approximated by \( n\alpha \) and \( n\beta \) respectively, minus some small fractional parts. Thus, the left-hand side (LHS) grows like:
   \[
   [n\alpha] \cdot [n\beta] \approx (n\alpha - \{n\alpha\})(n\beta - \{n\beta\}) = n^2\alpha\beta - n(\alpha\{n\beta\} + \beta\{n\alpha\}) + \{n\alpha\}\{n\beta\}
   \]
   Similarly, the right-hand side (RHS) grows like:
   \[
   [n\gamma] \cdot [n\delta] \approx n^2\gamma\delta - n(\gamma\{n\delta\} + \delta\{n\gamma\}) + \{n\gamma\}\{n\delta\}
   \]

3. **Equating Leading Terms**: Since the leading terms on both sides must be equal for all \( n \), we have:
   \[
   n^2\alpha\beta \approx n^2\gamma\delta
   \]
   Dividing both sides by \( n^2 \) and taking the limit as \( n \to \infty \), we get:
   \[
   \alpha\beta = \gamma\delta
   \]

4. **Analyzing the Error Terms**: Next, we consider the error terms. Using the floor function properties, we write:
   \[
   [n\alpha] = n\alpha - \{n\alpha\}, \quad [n\beta] = n\beta - \{n\beta\}
   \]
   Substituting these into the original equation, we get:
   \[
   (n\alpha - \{n\alpha\})(n\beta - \{n\beta\}) = (n\gamma - \{n\gamma\})(n\delta - \{n\delta\})
   \]
   Expanding and simplifying, we obtain:
   \[
   n^2\alpha\beta - n(\alpha\{n\beta\} + \beta\{n\alpha\}) + \{n\alpha\}\{n\beta\} = n^2\gamma\delta - n(\gamma\{n\delta\} + \delta\{n\gamma\}) + \{n\gamma\}\{n\delta\}
   \]

5. **Equating the Linear Terms**: The linear terms must also be equal:
   \[
   \alpha\{n\beta\} + \beta\{n\alpha\} = \gamma\{n\delta\} + \delta\{n\gamma\}
   \]
   Dividing by \( n \) and considering the average value over all \( n \), we use the fact that for irrational \(\alpha\), the fractional parts \(\{n\alpha\}\) are equidistributed modulo 1. Thus, the average value of \(\{n\alpha\}\) is \( \frac{1}{2} \), and similarly for the other terms. This gives:
   \[
   \frac{\alpha + \beta}{2} = \frac{\gamma + \delta}{2}
   \]
   Therefore:
   \[
   \alpha + \beta = \gamma + \delta
   \]

6. **Conclusion Using Vieta's Formulas**: We now have two equations:
   \[
   \alpha\beta = \gamma\delta \quad \text{and} \quad \alpha + \beta = \gamma + \delta
   \]
   These are the Vieta's formulas for the roots of a quadratic equation. Hence, the sets \(\{\alpha, \beta\}\) and \(\{\gamma, \delta\}\) must be equal.

\(\blacksquare\)

The final answer is \( \boxed{ \{\alpha, \beta\} = \{\gamma, \delta\} } \)