34. Solution 1. Let there be $N$ students in the school before the New Year, among which there were $M$ boys, constituting $k$ percent of the total. This means that $M = \frac{k}{100} N$, or $100 M = k N$.

After the New Year, the number of boys became $M+1$, and the total number of students became $N+3$. If now the boys constitute $\ell$ percent (by condition $\ell < 100$, since after the New Year there are definitely some girls in the school), then

$$
100(M+1) = \ell(N+3)
$$

Recalling the equality $100 M = k N$, we find that

$$
\ell N + 3 \ell = 100 M + 100 = k N + 100
$$

from which $100 - 3 \ell = (\ell - k) N$. If $3 \ell < 100$ (i.e., girls constitute less than a third), then the number $N$ is a natural divisor of the positive number $100 - 3 \ell < 100$, and therefore is less than 100. If, however, $3 \ell$ is greater than 100 (it obviously cannot be equal to 100), then $(k - \ell) N = 3 \ell - 100 \leqslant 3 \cdot 99 - 100 = 197$, and the number of students in the old year is no more than 197.
Solution 2. Let $m$ and $d$ be the number of boys and girls in the school before New Year. Let $\frac{100 m}{m+d}=x$, where $x$ is an integer by the problem's condition.

First, consider the case where the proportion of boys has not changed, i.e., $\frac{100(m+1)}{m+d+3}=x$. Then

$$
\frac{100 m}{m+d}=\frac{100(m+1)}{m+d+3}
$$

which implies $100 m(m+d+3)=100(m+1)(m+d)$, i.e., $3 m=m+d$. Thus, in this case, boys make up a third of the total number of students, which is not an integer percentage. Therefore, this case is impossible.

We are left to consider the cases

$$
\frac{100 m}{m+d}-\frac{100(m+1)}{m+d+3} \geqslant 1 \quad \text { and } \quad \frac{100(m+1)}{m+d+3}-\frac{100 m}{m+d} \geqslant 1
$$

These cases are analogous; let's analyze the first one. Multiplying by the denominators, we get the inequality

$$
100 m(m+d+3)-100(m+1)(m+d) \geqslant(m+d)(m+d+3)
$$

Expanding the brackets and transforming the left side, we get:

$$
100(2 m-d) \geqslant(m+d)(m+d+3)
$$

from which $m+d \leqslant \frac{100(2 m-d)}{m+d+3}$. Using the obvious inequality $\frac{2 m-d}{m+d+3}<2$, we obtain that $m+d<200$.