A man disposes of sufficiently many metal bars of length $2$ and wants to construct a grill of the shape of an $n \times n$ unit net. He is allowed to fold up two bars at an endpoint or to cut a bar into two equal pieces, but two bars may not overlap or intersect. What is the minimum number of pieces he must use?
To solve the problem, we need to determine the minimum number of metal pieces required to construct an \( n \times n \) unit net using metal bars of length 2. The key is to decompose the net into L-shaped pieces.

1. **Calculate the total length of metal required:**
   An \( n \times n \) unit net consists of \( n+1 \) horizontal lines and \( n+1 \) vertical lines, each of length \( n \). Therefore, the total length of metal required is:
   \[
   2n(n+1)
   \]
   This is because there are \( n+1 \) horizontal lines and \( n+1 \) vertical lines, each of length \( n \), so the total length is:
   \[
   (n+1)n + (n+1)n = 2n(n+1)
   \]

2. **Decompose the net into L-shaped pieces:**
   Each L-shaped piece consists of two segments of length 1, forming a right angle. Each L-shaped piece can be considered as using 2 units of metal length.

3. **Determine the number of L-shaped pieces:**
   Since each L-shaped piece uses 2 units of metal length, the number of L-shaped pieces required to cover the total length \( 2n(n+1) \) is:
   \[
   \frac{2n(n+1)}{2} = n(n+1)
   \]

4. **Verify the construction:**
   To ensure that the construction is valid, we need to check that the L-shaped pieces can be arranged without overlapping or intersecting. This can be done by visualizing the net and placing the L-shaped pieces accordingly. For example, for \( n=3 \), the net can be decomposed into L-shaped pieces as shown in the problem statement.

Therefore, the minimum number of pieces required to construct the \( n \times n \) unit net is \( n(n+1) \).

The final answer is \( \boxed{ n(n+1) } \).