On the evening, more than $\frac 13$ of the students of a school are going to the cinema. On the same evening, More than $\frac {3}{10}$ are going to the theatre, and more than $\frac {4}{11}$ are going to the concert. At least how many students are there in this school?
1. **Define Variables and Total Students**:
   Let \( a \) be the number of students going to the cinema, \( b \) be the number of students going to the theatre, and \( c \) be the number of students going to the concert. Let \( S \) be the total number of students in the school. We are given the following inequalities:
   \[
   \frac{a}{S} > \frac{1}{3}, \quad \frac{b}{S} > \frac{3}{10}, \quad \frac{c}{S} > \frac{4}{11}
   \]
   which can be rewritten as:
   \[
   3a > S, \quad 10b > 3S, \quad 11c > 4S
   \]

2. **Express Total Students in Terms of \( a, b, c \)**:
   Suppose \( S = a + b + c \). We need to find \( a, b, c \in \mathbb{Z} \) such that the inequalities hold simultaneously. 

3. **Initial Assumption and Inequalities**:
   Assume \( 2a = b + c + k \) where \( k = 1 \). We will later consider the possibilities when \( k > 1 \). The inequalities become:
   \[
   2a > b + c
   \]
   \[
   7b > \frac{3}{2}(b + c + k) + 3c \implies 11b > 3k + 9c
   \]
   \[
   7c > 2(b + c + k) + 4b \implies 5c > 6b + 2k
   \]

4. **Combining Inequalities**:
   Combining the inequalities, we get:
   \[
   11b > 3k + \frac{9}{5}(6b + 2k) \implies b > 33k
   \]
   where \( k \) is some positive integer.

5. **Establishing Bounds on \( c \)**:
   Using \( 2a = b + c + 1 \), we establish upper and lower bounds on \( c \) in terms of \( b \) and \( k \):
   \[
   7b > \frac{3}{2}(b + c + 1) + 3c \implies 14b > 3b + 9c + 3 \implies \frac{11b - 3}{9} > c
   \]
   \[
   7c > 2(b + c + 1) + 4b \implies 5c > 6b + 2 \implies c > \frac{6b + 2}{5}
   \]

6. **Finding Interval for \( c \)**:
   We need the interval:
   \[
   \left( \frac{6b + 2}{5}, \frac{11b - 3}{9} \right)
   \]
   to contain an integer. Consider the endpoints modulo 1. If the fractional part of the right endpoint is less than the fractional part of the left endpoint, an integer must be contained in the interval.

7. **Simplifying the Interval**:
   Using the substitution \( b' = b - 33 \) for ease of calculations, we know \( b \geq 34 \) and hence \( b' \geq 1 \). This changes the interval to:
   \[
   \left( \frac{b' + 35}{5}, \frac{2b' + 63}{9} \right) \equiv \left( \frac{b'}{5}, \frac{2b'}{9} \right)
   \]
   Using a common denominator, we get:
   \[
   \left( \frac{9b'}{45}, \frac{10b'}{45} \right)
   \]

8. **Finding the Smallest \( b' \)**:
   We need the smallest \( b' \) such that the fractional part of the left endpoint is greater than the fractional part of the right endpoint. We begin computations from \( b' = 1 \):
   \[
   (9, 10), (18, 20), (27, 30), (36, 40), (0, 5), (9, 15)
   \]
   \[
   (18, 25), (27, 35), (36, 0), (0, 10), (9, 20), (18, 30)
   \]
   \[
   (27, 40), \boxed{(36, 5)}, (0, 15), (9, 25), (18, 35), (27, 0), \boxed{(36, 10)}
   \]

9. **Finding the Minimum Sum**:
   The first boxed pair corresponds to \( b' = 14, b = 47 \) and the interval \( \left( \frac{284}{5}, \frac{514}{9} \right) \). This interval contains \( 57 \), but \( 2a = b + c + 1 = 105 \) implies \( a \) is not an integer. The second boxed pair corresponds to \( b' = 19, b = 52 \) and the interval \( \left( \frac{314}{5}, \frac{569}{9} \right) \), which contains \( 63 \). Now, \( 2a = 116 \implies a = 58 \), and the minimum sum when \( k = 1 \) is \( 19 + 52 + 58 = 173 \).

10. **Considering \( k \geq 2 \)**:
    If \( k \geq 2 \), then \( b > 66 \), and it is not possible for \( k \geq 2 \) to yield a triplet \( (a, b, c) \) with a sum less than what we have already constructed.

Hence, the minimum number of possible students is \( \boxed{173} \).