Given a complete bipartite graph on $n,n$ vertices (call this $K_{n,n}$), we colour all its edges with $2$ colours , red and blue . What is the least value of $n$ such that for any colouring of the edges of the graph , there will exist at least one monochromatic $4$ cycle ?
1. **Step 1: Verify the case for \( n = 4 \)**

   We need to show that there exists a coloring of the edges of \( K_{4,4} \) that avoids any monochromatic \( C_4 \) (4-cycle). Consider the following coloring:
   
   - Vertices on the left side: \( \{0, 1, 2, 3\} \)
   - Vertices on the right side: \( \{0, 1, 2, 3\} \)
   
   Connect vertex \( i \) on the left to:
   - Vertices \( i \) and \( i+1 \) (mod 4) on the right with blue edges.
   - Vertices \( i+2 \) and \( i+3 \) (mod 4) on the right with red edges.
   
   This coloring avoids any monochromatic \( C_4 \). Therefore, \( n = 4 \) does not guarantee a monochromatic \( C_4 \).

2. **Step 2: Prove the case for \( n = 5 \)**

   We need to show that for any coloring of the edges of \( K_{5,5} \), there will always be a monochromatic \( C_4 \). Enumerate the vertices of one partite set as \( \{1, 2, 3, 4, 5\} \). Let \( r_i \) be the number of red edges from vertex \( i \) and \( b_i \) be the number of blue edges from vertex \( i \). Note that \( r_i + b_i = 5 \) for each \( i \).

3. **Step 3: Count the number of monochromatic \( K_{2,2} \)**

   A \( 4 \)-cycle is just a \( K_{2,2} \). If two vertices are connected with red (or blue) edges to two other vertices, we have a monochromatic \( K_{2,2} \). The number of such triples \( (u, v, w) \) where \( w \) is from the chosen partite set and \( u, v \) are from the other partite set such that both edges \( uw \) and \( vw \) are red is at most \( \binom{5}{2} = 10 \).

4. **Step 4: Use combinatorial arguments**

   The number of such triples is exactly \( \sum_{i} \binom{r_i}{2} \). Therefore, we have:
   \[
   \binom{5}{2} \geq \sum_{i} \binom{r_i}{2}
   \]
   Similarly, for blue edges:
   \[
   \binom{5}{2} \geq \sum_{i} \binom{b_i}{2}
   \]

5. **Step 5: Apply convexity**

   Since \( r_i + b_i = 5 \), by convexity, we have:
   \[
   \binom{r_i}{2} + \binom{b_i}{2} \geq 4
   \]
   with equality only when \( r_i \in \{2, 3\} \).

6. **Step 6: Sum the inequalities**

   Adding the two bounds obtained earlier:
   \[
   2 \binom{5}{2} = 5 \cdot 4 = 20
   \]
   This implies:
   \[
   \binom{5}{2} = \sum_{i} \binom{r_i}{2}
   \]
   and each \( r_i \in \{2, 3\} \).

7. **Step 7: Check for contradiction**

   If \( r_i \in \{2, 3\} \), then \( \binom{r_i}{2} \) is odd. Summing five odd numbers should yield an odd number, but \( 10 = \binom{5}{2} \) is even. This contradiction implies that no matter how we color the edges of \( K_{5,5} \), some monochromatic \( C_4 \) will be induced.

\(\blacksquare\)

The final answer is \( \boxed{ n = 5 } \)