As $E$ is a point on the segment $AC$, therefore $CE=n \leq AC=21$, and from the (possibly degenerate) isosceles triangle $ADE$, $AD+DE \geq AE$, which means $n \geq 7$.

If now $n=21$, then $E$ and $A$ coincide, $D$ is the point on the segment $AB$ such that $AD=21$, and $BC=m$ can be any integer that allows the formation of the triangle $ABC$. Thus, one possible value of $n$ is 21.

$$
\text { 1985-05-213-1.eps }
$$

If $n=7$, then $E$ is the point closer to $C$ that divides the segment $AC$ into a ratio of 1:2. The only point that is $n=7$ units away from both $A$ and $E$ is the midpoint of $AE$. Therefore, $D$ must coincide with this point, but then it cannot lie on the side $AB$ of the triangle $ABC$. Thus, $n \neq 7$, or $n>7$.

In the remaining cases where $7<n<21$, $ADE$ is a true isosceles triangle. Let the perpendicular projections of $B$ and $D$ onto $AC$ be $B'$ and $D'$. Then $AD' = AE / 2 = (21-n) / 2$. The right triangles $AD'D$ and $AB'B$ are similar, with the ratio of similarity being $AB: AD = 33: n$, so

$$
AB' = AD' \cdot \frac{AB}{AD} = \frac{21-n}{2} \cdot \frac{33}{n}
$$

Now $B'$ can fall either on the side $AC$ or on its extension beyond $C$, in any case $CB' = |AC - AB'|$. For the right triangles $AB'B$ and $CB'B$, by the Pythagorean theorem,

$$
AB^2 - AB'^2 = BB'^2 = BC^2 - CB'^2
$$

Substituting $AB=33$, $BC=m$, and $CB' = |21 - AB'|$,

$$
33^2 - AB'^2 = m^2 - 21^2 + 2 \cdot 21 \cdot AB' - AB'^2
$$

From (1), rearranging,

$$
n(2223 - m^2) = 3^3 \cdot 7^2 \cdot 11
$$

The second factor on the left side is an integer, so $n$ must be a divisor of the right side. We also know that $7 < n < 21$, so the possible values for $n$ are $n=9$ and $n=11$. In the first case, $m^2 = 2223 - 3 \cdot 7^2 \cdot 11 = 606$, which does not yield an integer value for $m$. However, if $n=11$, then $m^2 = 2223 - 3^3 \cdot 7^2 = 900$, so $m=30$.

Thus, in addition to the value $n=21$, only $n=11$ is possible. We still need to verify that in the triangle $ABC$ where $BC=m=30$ units, suitable points $D$ and $E$ can indeed be found. By reversing the steps of the above reasoning, this is easily confirmed, so we will not detail it.

In summary, the possible values of $n$ are 11 and 21.
In triangle $ABC$, $AB=33$, $AC=21$, and $BC=m$ units, where $m$ is an integer. Find points $D$ on $AB$ and $E$ on $AC$ such that $AD=DE=EC=n$, where $n$ is also an integer. What can the value of $n$ be?