Christmas Streak 01/88: EINS + EINS + EINS + EINS = VIER!

$1+1+1+1=4,~2+2+2+2=8,~3+3+3+3=12,~4+4+4+4=16,~5+5+5+5=20, \\ \\ 6+6+6+6=24,~7+7+7+7=28,~8+8+8+8=32,~9+9+9+9=36.$

Since the carryover cannot be bigger than $3,$ and from $I+I+I+I+\text{carryover}_{\text{tenth}}\equiv I \pmod{10},$

$I\neq 1,~2,~4,~5,~7,~8~\Rightarrow~I=0,~3,~6,~9.$

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$\large {\begin{array}{cccccc} && &\color{#D61F06}2&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ && &\color{#D61F06}2&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ && &\color{#D61F06}2&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ +&& &\color{#D61F06}2&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ \hline && &\color{#BA33D6}V&\color{#EC7300}I&\color{#D61F06}2&\color{#E81990}R \end{array}}$

$E=1~\text{or}~2.$

If $E=2,$ then the carryover from the hundredth digit cannot be bigger than $1.$

Since $I\neq1,~2,~4,$ the only possible values for $I$ are $0$ or $3.$

$I=0)$ Then, the carryover from the tenth digit must be 0. $N=1$ doesn't satisfy the tenth-digit condition, and $N=0,~2$ is impossible, therefore $I\neq0.$

$I=3)$ Then, the carryover from the tenth digit must be 1. $N=4$ doesn't satisfy the tenth-digit condition, and $N=3$ is impossible, therefore $I\neq3.$

This is a contradiction, therefore, $E=1.$

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$\large {\begin{array}{cccccc} && &\color{#D61F06}1&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ && &\color{#D61F06}1&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ && &\color{#D61F06}1&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ +&& &\color{#D61F06}1&\color{#EC7300}I&\color{#20A900}N&\color{#3D99F6}S \\ \hline && &\color{#BA33D6}V&\color{#EC7300}I&\color{#D61F06}1&\color{#E81990}R \end{array}}$

If $I=0,$ then the carryover from the tenth digit is 0. This is impossible because $N$ needs to be 1 or 2, but the tenth-digit condition cannot be satisfied.

If $I=6,$ then the carryover from the hundredth digit is 2. And then $V=6.$ Nope,

If $I=9,$ then the carryover from the hundredth digit is 3 $(\Rightarrow~V=7)$ and the carryover from the tenth digit is also 3. This is impossible because $N$ needs to be 8 or 9, but the tenth-digit condition cannot be satisfied.

Therefore, $I=3.$ Then, $V=5.$

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$\large {\begin{array}{cccccc} && &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}N&\color{#3D99F6}S \\ && &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}N&\color{#3D99F6}S \\ && &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}N&\color{#3D99F6}S \\ +&& &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}N&\color{#3D99F6}S \\ \hline && &\color{#BA33D6}5&\color{#EC7300}3&\color{#D61F06}1&\color{#E81990}R \end{array}}$

$N=4$ doesn't satisfy the tenth-digit condition, whereas $N=3$ is prohibited by $I=3.$

So we figure that $N=2.$ Then $S$ needs to create a carryover of 3. $S=8$ would create $R=2,$ which is prohibited by $N=2.$

So, $S=9,$ and $R=6.$

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$\large {\begin{array}{cccccc} && &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}2&\color{#3D99F6}9 \\ && &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}2&\color{#3D99F6}9 \\ && &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}2&\color{#3D99F6}9 \\ +&& &\color{#D61F06}1&\color{#EC7300}3&\color{#20A900}2&\color{#3D99F6}9 \\ \hline && &\color{#BA33D6}5&\color{#EC7300}3&\color{#D61F06}1&\color{#E81990}6 \end{array}}$

Therefore, the answer is $\boxed{5316}.$