Solve the equation $n+S(n)=1981$, where $S(n)$ denotes the sum of the digits of the natural number $n$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
I. solution. Since $S(n) \geq 0$, thus $n \leq 1981$, so $S(n)1981-28=1953$. This, however, means that $S(n) \geq 1+9+6=16$, so $n \leq 1981-16=1965$.

It is known that a number written in base 10, when divided by 9, gives the same remainder as the sum of its digits. Since the remainder of 1981 when divided by 9 is 1, $n$ and $S(n)$, when divided by 9, can only give a remainder of 5. There is only one such number between 1954 and 1965, which is 1958. For this, $S(1958)=23$, and $1958+23=1981$, so this is the only solution to the problem.

II. solution. Since $n \leq 1981$, $n$ can have at most four digits, so it can be written in the form $1000 x+100 y+10 z+v$, where $x, y, z$ and $v$ are natural numbers between 0 and 9, the digits of $n$. Then $S(n)=x+y+z+v$.

According to the condition,

$$
1001 x+101 y+11 z+2 v=1981
$$

Since $0 \leq 101 y+11 z+2 v \leq 101 \cdot 9+11 \cdot 9+2 \cdot 9=1026$, from (1) we get $1981-1026=955 \leq 1001 x \leq 1981$, so

$$
0<\frac{955}{1001} \leq x \leq \frac{1981}{1001}<2
$$

The obtained condition is satisfied only by $x=1$ among the natural numbers. Substituting this into (1):

$$
101 y+11 z+2 v=980
$$

Since $0 \leq 11 z+2 v \leq 11 \cdot 9+2 \cdot 9=117$, from (2) we get $980-117=863 \leq 101 y \leq 980$ so,

$$
8<\frac{893}{101} \leq y \leq \frac{980}{101}<10
$$

The obtained condition is satisfied only by $y=9$ among the natural numbers. Substituting this into (2):

$$
11 z+2 v=71
$$

Since $0 \leq 2 v \leq 2 \cdot 9=18$, from (3) we get $71-18=53 \leq 11 z \leq 71$, so

$$
4<\frac{53}{11} \leq z \leq \frac{71}{11}<7
$$

The obtained condition is satisfied by both $z=5$ and $z=6$ natural numbers, but in the latter case $2 v=71-66=5$ would be, which is not possible because $v$ is an integer. If $z=5$, then $2 v=71-55=16$, so $v=8$.

Thus $x=1, y=9, z=5, v=8$, so $n=1958$.