For a fixed positive integer $n \geq 3$ we are given a $n$ $\times$ $n$ board with all unit squares initially white. We define a [i]floating plus [/i]as a $5$-tuple $(M,L,R,A,B)$ of unit squares such that $L$ is in the same row and left of $M$, $R$ is in the same row and right of $M$, $A$ is in the same column and above $M$ and $B$ is in the same column and below $M$. It is possible for $M$ to form a floating plus with unit squares that are not next to it. Find the largest positive integer $k$ (depending on $n$) such that we can color some $k$ squares black in such a way that there is no black colored floating plus.

[i]Proposed by Nikola Velov[/i]
1. **Define the Problem and Initial Setup:**
   We are given an \( n \times n \) board with all unit squares initially white. A *floating plus* is defined as a 5-tuple \((M, L, R, A, B)\) of unit squares such that:
   - \(L\) is in the same row and to the left of \(M\),
   - \(R\) is in the same row and to the right of \(M\),
   - \(A\) is in the same column and above \(M\),
   - \(B\) is in the same column and below \(M\).

   We need to find the largest positive integer \(k\) such that we can color some \(k\) squares black in such a way that there is no black colored floating plus.

2. **Prove that \(4n-4\) is Possible:**
   - Color all the unit squares on the edges of the board black. There are \(n\) unit squares on each edge, and the corners are counted twice, so this gives us exactly \(4n-4\) black squares.
   - To see that they form no black floating plus, note that the center \(M\) of any floating plus lies in a row with at least three black squares and in a column with at least three black squares. In this coloring, the only possible squares for \(M\) are thus the corners of the board, but obviously, they cannot form a floating plus.

3. **Prove that \(4n-4\) is the Maximum:**
   - Assume that there are at least \(4n-3\) black squares and no black floating plus is formed.
   - Every black unit square is not the center \(M\) of a black floating plus. This means that at least one of the four directions \(\leftarrow\), \(\rightarrow\), \(\uparrow\), and \(\downarrow\) with respect to that square does not contain another black unit square.
   - Consider all pairs of the form \((s, \leftarrow)\), \((s, \rightarrow)\), \((s, \uparrow)\), and \((s, \downarrow)\) where \(s\) is a black square and the arrow points in the direction with respect to \(s\) that has no other black square. Let there be \(P\) such pairs. Because no black square is the center of a black floating plus, there are at least \(4n-3\) such pairs.

4. **Analyze the Pairs:**
   - Observe the leftmost column that contains black squares (all columns to the left of it are entirely white). The top black unit square in this column \(x\) is part of at least two pairs, \((x, \leftarrow)\) and \((x, \uparrow)\). The bottom black unit square in this column \(y\) is part of at least two pairs \((y, \leftarrow)\) and \((y, \downarrow)\). This means that the black unit squares in this column generate at least two extra pairs, even if \(x=y\).
   - The same is true for the rightmost column. Now notice that the leftmost and the rightmost column are not the same, because that would imply that we have at most \(n\) squares, which is strictly less than \(4n-3\). This means that we have at least \((4n-3)+4 = 4n+1\) pairs, in other words, \(P \geq 4n+1\).

5. **Apply the Pigeonhole Principle:**
   - Every pair has four possible directions for the arrow. It follows from \(P \geq 4n+1\) and the pigeonhole principle that there are at least \(n+1\) pairs with the same direction, for example, the direction \(\rightarrow\).
   - Thus, we have some pairs \((s_{1}, \rightarrow)\), \((s_{2}, \rightarrow)\), ..., \((s_{n+1}, \rightarrow)\). However, there are only \(n\) rows, so there must be two black squares \(s_{i}\) and \(s_{j}\) in the same row. This means that the pairs \((s_{i}, \rightarrow)\) and \((s_{j}, \rightarrow)\) are from the same row, which is not possible, because one is to the right of the other. This contradiction shows that there are at most \(4n-4\) black squares.

\(\blacksquare\)

The final answer is \( \boxed{ 4n-4 } \)