The ratio of the areas of the equilateral triangle $\Delta$ of side $n + \frac{1}{2n}$ and that of the equilateral triangle $\Delta_1$ of side $1$ is the square of the ratio of the lengths of their sides, i.e.
$$(n + \frac{1}{2n})^2 > n^2 + 1,$$
hence at least $n^2 + 2$ triangles $\Delta_1$ are needed.

For $n = 1$ we can do that by placing three $\Delta_1$ triangles at the corners.

Assume now this proven until $n$, and prove by induction for $n+1$. A $\Delta$ of side $n+1$ triangle placed at the top corner will use $n^2 + 2$ triangles $\Delta_1$, according with the induction hypothesis. It remains a trapezoidal strip at the bottom, of length of the nonparallel sides
$$n + 1 + \frac{1}{2(n+1)} - n - \frac{1}{2n} = 1 - \frac{1}{2n(n+1)},$$
and basis lengths $n + \frac{1}{2n}$ and $n + 1 + \frac{1}{2(n+1)}$, with $(n+1)^2 + 2 - n^2 - 2 = 2n + 1$ triangles $\Delta_1$ available to cover it.

Place $2n+1$ triangles $\Delta_1$ one next to another, every second one "slid" downwards by $\frac{1}{2n(n+1)}$. They will cover a trapezoidal strip of exactly the dimensions of the above, since
$$(n+1) \cdot 1 + n \cdot \frac{1}{2n(n+1)} = n + 1 + \frac{1}{2(n+1)}.$$