Belarusian Mathematical Olympiad

Answer: $60^\circ$.

Let $E$ and $F$ be the midpoints of the segments $BM$ and $CM$, respectively. Since $\angle AMB = \angle CMD = 60^\circ$, we have
$$\begin{aligned} \angle BMC &= 360^\circ - \angle AMB - \angle CMD - \angle LMN = 360^\circ - 60^\circ - 60^\circ - \angle LMN = \\ &= 240^\circ - \angle LMN. \end{aligned}$$
Since $KF$ is the midline in the triangle $CBM$, we have $KF \parallel BM$, therefore, $\angle KFC = \angle BMC$. Then
$$\begin{aligned} \angle KFM &= 180^\circ - \angle KFC = 180^\circ - \angle BMC = \\ &= 180^\circ - (240^\circ - \angle LMN) = \angle LMN - 60^\circ. \end{aligned}$$
By condition, $CM = DM$ and $\angle CMD = 60^\circ$, so the triangle $CMD$ is equilateral. Since $FN$ is the midline in the triangle $CMD$, we see that the triangle $FMN$ is equilateral too and $FM = NM = FN$, $\angle MFN = 60^\circ$. Therefore,
$$\angle KFN = \angle KFM + \angle MFN = [\angle MFN = 60^\circ] = \angle LMN - 60^\circ + 60^\circ = \angle LMN.$$
In the same way, one can show that $\angle KEL = \angle LMN$ and $EM = ML = EL$. Thus, $\triangle KEL = \triangle NFK = \triangle NML$ (by two sides and the angle between them), so $KL = KN = LN$. Therefore, the triangle $KLN$ is equilateral and $\angle LKN = 60^\circ$.