Let $n$ be an even positive integer, representing the dimensions of an $n \times n$ board. We need to determine the minimal number of cells that must be marked on the board such that every cell, whether marked or unmarked, has at least one marked neighboring cell.

A cell on the board has neighboring cells that share a common side. For a cell located at position $(i, j)$, its potential neighbors are located at $(i-1, j)$, $(i+1, j)$, $(i, j-1)$, and $(i, j+1)$, provided these indices remain within the bounds of the board.

### Plan

To cover all cells with the minimum number of marked cells, an efficient strategy is to mark cells in a checkerboard pattern. This strategy ensures that at least one neighbor of each non-marked cell is marked since the checkerboard pattern alternates between marked and unmarked cells.

### Calculation

The checkerboard pattern results in two possible configurations, each of which ensures that half of the cells on the board are marked. The minimal marking is achieved as follows:

1. Since $n$ is even, we can split the board into two equal halves: cells are alternately marked on checkerboard patterns across the $n$ rows.
2. This arrangement leads to exactly half of the total number of cells being marked.

The total number of cells on the board is $n^2$. In a checkerboard pattern:
- Each row has $\frac{n}{2}$ cells marked.
- There are $n$ such rows.

Thus, the total number of marked cells necessary is:
$$\left( \frac{n}{2} \right) \times n = \frac{n^2}{2}.$$

We further optimize this by observing that marking every alternate row reduces the number of marked rows by half:
- Only $\frac{n}{2}$ rows need to be entirely marked.

Therefore, we adjust the board:
- Mark $\frac{n}{2}$ rows, each with $\frac{n}{2} + 1$ marked cells due to boundary considerations.

The number of marked cells in these rows can be calculated as:
$$\left( \frac{n}{2} \right) \times \left( \frac{n}{2} + 1 \right) = \frac{n^2}{4} + \frac{n}{2}.$$

Thus, the minimal number of cells required to be marked on an $n \times n$ board, where $n$ is even, ensuring that every cell has a marked neighbor is:
$$\boxed{\frac{n^2}{4} + \frac{n}{2}}.$$