(I)    ISE + EID  = ALK
        -      +     -
(II)   RAS -  KL  = RUM
       ----------------
(III)  ERR + MDR  = LLM
       (IV)   (V)   (VI)


Solution:
(VI.2) shows U = 0 or U = 9. Now no other letter can be zero.
As there are no leading digits I, E, A, R, K, M, L cannot be zero.
S is not zero from (IV.1). D is not zero from (I.1).
Therefore U = 0.
No other letter can be recogniced directly.
Assume R is known then
(III.1)  M == 2R
(VI.1)   K == 2M    == 4R
(V.3)    E == M - 1 == 2R - 1  (with carry)
(I.1)    D == K - E == 2R + 1
(IV.1)   S == E - R ==  R - 1
(II.1)   L == S - M == -R - 1
(VI.3)   A == L + R == -1      (no carry)
Therefore A = 9.
(II.2)   K = A - 1 = 8.
As K == 4R (mod 10) we can conclude 3 = 4R (mod 5) that is 3 + R = 0 (mod 5).
Therefore R must be 2 or 7. As L > E from (III.3) only R = 2 is possible.
Than M = 4, E = 3, D = 5, S = 1, L = 7.
The last letter I = 6.

       613 + 365  = 978
        -      +     -
       291 -  87  = 204
       ----------------
       322 + 452  = 774


References:

H. Müller-Merbach;
  The Role of Puzzles in Teaching Combinatorial Programming,
  In: B. Roy (ed)., Combinatorial Programming: Methods and 
  Applications, 379-386, D. Reidel Publ. 1975

Zeit Magazin 19/1972