3. Given a convex $n$-sided polygon, each of its diagonals is the perpendicular bisector of at least one other diagonal. Find all possible positive integers $n(n>3)$.

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3. Suppose there exists an $n$-sided polygon satisfying the given conditions. Let the set of all its diagonals be $T$.
From the given, we can construct a mapping
$$
f: T \rightarrow T, l \mapsto m=f(l),
$$

where $l$ is the perpendicular bisector of $m$.
If such an $m$ is not unique, then choose any one.
If there exist $l_{1}, l_{2}$ such that $f(l_{1})=f(l_{2})=m$, then $l_{1}, l_{2}$ are both perpendicular bisectors of $m$. Thus, $l_{1}=l_{2}$. This shows that $f$ is injective. Since $T$ is a finite set, $f$ is a bijective mapping. Therefore, for each diagonal, there must exist a unique other diagonal that is its perpendicular bisector.

As shown in Figure 19, let $B, A, C$ be three consecutive vertices.

Considering the diagonal $BC$, only the diagonals passing through point $A$ can intersect it. Therefore, the perpendicular bisector of $BC$ must pass through point $A$. Thus, $AB=AC$.

By the generality of the selection of points $B, A, C$, we know that all sides of the $n$-sided polygon are equal.

As shown in Figure 20, let $A, B, C, D$ be four consecutive vertices.

If $n>4$, then $AD$ is a diagonal, and its perpendicular bisector must pass through point $B$ or $C$. Without loss of generality, assume the perpendicular bisector of $AD$ passes through point $B$. Thus, $BD=BA$.

Since $AB=BC=CD$, $\triangle BCD$ is an equilateral triangle, $\angle C=60^{\circ}$. This indicates that in any two consecutive interior angles of this polygon, one must be $60^{\circ}$. Therefore, the polygon has at least $\frac{n}{2}$ interior angles equal to $60^{\circ}$.

By a well-known result, when $n>3$, any convex polygon has at most two interior angles equal to $60^{\circ}$. Thus, $\frac{n}{2} \leqslant 2$. Therefore, $n \leqslant 4$, leading to a contradiction.
Clearly, any rhombus satisfies the given conditions.
Hence, $n=4$ is the required value.