Solution:

To hold equal charges \(q\) in equilibrium at the corners of a square, we need to place a charge \(Q\) at the center of the square such that the net force on each corner charge due to other charges is balanced.

1. Force due to corner charges:
Each charge \(q\) at the corners experiences repulsive forces due to the other three corner charges. The net force from these three charges is:
\[
F_{\text{corners}} = q \cdot \frac{1}{4\pi\varepsilon_0} \left( \frac{2}{a^2} + \frac{\sqrt{2}}{a^2} \right) = \frac{q^2}{4\pi\varepsilon_0 a^2} \left( 2 + \sqrt{2} \right),
\]
where \(a\) is the side of the square.

2. Force due to center charge \(Q\):
The attractive force due to the central charge \(Q\) on each corner charge is:
\[
F_{\text{center}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{r^2},
\]
where \(r = \frac{a}{\sqrt{2}}\) is the distance from the center to a corner. Substituting \(r\):
\[
F_{\text{center}} = \frac{qQ}{4\pi\varepsilon_0 \cdot \frac{a^2}{2}} = \frac{2qQ}{4\pi\varepsilon_0 a^2}.
\]

3. Equilibrium condition:
For equilibrium, \(F_{\text{corners}} = F_{\text{center}}\):
\[
\frac{q^2}{4\pi\varepsilon_0 a^2} \left( 2 + \sqrt{2} \right) = \frac{2qQ}{4\pi\varepsilon_0 a^2}.
\]

4. Solve for \(Q\):
\[
Q = \frac{q}{2} \left( 2 + \sqrt{2} \right).
\]
Since \(Q\) is opposite in sign to \(q\), the final charge is:
\[
Q = -\frac{q}{4} \left( 1 + 2\sqrt{2} \right).
\]

Thus, the required charge at the center is:

\[
Q = -\frac{q}{4} \left( 1 + 2\sqrt{2} \right).
\]