2. In the given expression, we remove the parentheses and group the numbers into quadruples:

$$
\underbrace{1-2-3+5}+\underbrace{6-7-8+10}+\underbrace{11-12-13+15}+\underbrace{16-17-18+20}+\ldots
$$

(the last group is shorter if the total number of numbers $N$ is not a multiple of four). The numbers in the $i$-th group $(i=1,2, \ldots)$ form the expression $(5 i-4)-(5 i-3)-(5 i-2)+5 i$, whose value is clearly equal to one (independent of the index $i$). Therefore, the value $V$ of the entire expression in the case of $N=4 k$ is $V=k$, in the case of $N=4 k+1$ is $V=k+(5 k+1)=6 k+1$, in the case of $N=4 k+2$ is $V=k+(5 k+1)-(5 k+2)=k-1$, and in the case of $N=4 k+3$ is $V=k+(5 k+1)-(5 k+2)-(5 k+3)=-4 k-4$. It is easy to determine when $V=103$:

$$
\begin{aligned}
& N=4 k, \quad V=k=103, \quad N=412 \\
& N=4 k+1, \quad V=6 k+1=103, \quad k=17, \quad N=69 \\
& N=4 k+2, \quad V=k-1=103, \quad k=104, \quad N=418 \\
& N=4 k+3, \quad V=-4 k-4=103, \quad k \notin \mathbb{N} .
\end{aligned}
$$

Answer: In Milan's expression, there were either 69, or 412, or 418 numbers.

For a complete solution, 6 points are awarded, and for correctly removing the parentheses, 2 points are awarded.
2. Milan wrote down several of the first natural numbers, omitting only the numbers 4, $9,14,19,24,29, \ldots$ Then he inserted the minus and plus signs alternately between the written numbers, so he got the expression

$$
1-2+3-5+6-7+8-10+11-12+13-15+\ldots
$$

Finally, he inserted a left parenthesis after each minus sign and the same number of right parentheses at the end of the expression:

$$
1-(2+3-(5+6-(7+8-(10+11-(12+13-(15+\ldots))))))
$$

The resulting expression had a value of 103. How many numbers were in Milan's expression? (Find all possibilities.)