Let's replace the letters with digits so that the following addition

| ÉLJEN  MÁJUS  ELSEJEJE |
| :---: |

is correct and at the same time, the sum of the digits in the two addends is equal. (Different letters, including E and É, represent different digits.)
For two 5-digit addends, the first digit of their 6-digit sum can only be 1, so $E=1$.

If $E=1$, then $N+S$ can only be 11.

$1+U+E$ also leaves a remainder, because $J+J=2J$ is even. Therefore, $J+J+1=1$, or 11, meaning $J=0$ or 5; but $1+U+E=10+J$, from which $U=8+J$, so only $J=0$ is possible (because $U$ cannot be 13) and thus $U=8$.

For the remaining 6 unknowns, the following 4 equations hold, where the unknowns can only be different positive single-digit numbers, excluding 0, 1, and 8.

$$
\begin{gathered}
N+S=11 \\
L+\bar{A}=S \\
\dot{E}+M=10+L \\
\dot{E}+L+1+N=M+\bar{A}+8+S
\end{gathered}
$$

which simplifies to

$$
\bar{E}+L+N=M+\bar{A}+S+7
$$

(2) cannot be replaced by $L+\bar{A}=10+S\left(2^{\prime}\right)$. Because in that case, (3) would become $\dot{E}+M+1=10+L$ (3'), and the sum of (1), (2'), and (3') would be $N+\bar{A}+\dot{E}+M=30$ after rearrangement, which is impossible because the left side can be at most $9+7+6+5=27$. From (2) and (3),

$$
S=\bar{E}+M-10+\bar{A}
$$

Substituting this value of $S$ into (1),

$$
N+\dot{E}+M+\hat{A}=21
$$

Substituting the values of $L$ and $S$ from (3) and (5) into (4) and rearranging,

$$
\dot{E}+N-M-2 \bar{A}=7
$$

Subtracting (7) from (6),

$$
2 M+3 \bar{A}=14
$$

Since $M=1$ is not possible, $-3 \bar{A}$ can only be 6, meaning

$$
\bar{A}=2 \quad \text { and } \quad M=4
$$

Substituting this value of $M$ into (3),

$$
\bar{E}-L=6
$$

Since $L$ is at least 3 and $\bar{E}$ is at most 9, only

$$
\bar{E}=9 \quad \text { and } \quad L=3 \text { are valid. }
$$

Substituting the values of $\bar{E}$, $M$, and $\bar{A}$ into (5) and (6), we get

$$
S=5, \quad N=6
$$

Thus, the solution is

$$
\begin{gathered}
93016+ \\
42085 \\
\hline 135101
\end{gathered}
$$

Quittner Pál (Bp., I., Petőfi g. II. o. t.)