A school has less than $5000$ students and if the students are divided equally into teams of either $9$ or $10$ or $12$ or $25$ each, exactly $4$ are always left out. However, if they are divided into teams of $11$ each, no one is left out. The maximum number of teams of $12$ each that can be formed out of the students in the school is

Let's call the number of students $N$. From the given conditions:

When divided by 9, 10, 12, or 25: remainder is 4

When divided by 11: remainder is 0 (no one left out)

$N < 5000$

---

When $N$ is divided by 9, 10, 12, or 25, the remainder is always 4. This means that if the actual number was 4 less, it would be perfectly divisible by 9, 10, 12, and 25

$(N - 4)$ is divisible by all of 9, 10, 12, and 25.

---

Let's find the smallest number that divides 9, 10, 12, and 25.

Prime factorization:

$9 = 3^2$

$10 = 2 \times 5$

$12 = 2^2 \times 3$

$25 = 5^2$

Taking the highest power of each prime factor:

$\text{LCM}(9, 10, 12, 25) = 2^2 \times 3^2 \times 5^2$

$= 4 \times 9 \times 25 = 900$

---

So $N - 4 = 900p$ for some integer $p$.

Therefore: $N = 900p + 4$

Now, we need to find a p such that N is divisible by 11 & N < 5000

Since, we know N < 5000, p would be < 5.

As maximum value of p is small, we can find the exact value by trial and error.

---

$p = 1$

$N = 900 (1) + 4$

$N = 904$

904 is not divisible by 11

$p = 2$

$N = 900 (2) + 4$

$N = 1804$

1804 is divisible by 11

Since we need to find the maximum value, we should check with the remaining too.

When $p = 3$, $N = 2704$ (Not divisible)

When $p = 4$, $N = 3604$ (Not divisible)

When $p = 5$, $N = 4504$ (Not divisible)

---

Max number of teams of 12 that can be formed:

$\dfrac{N}{12} = 150.33$

Hence, the answer is 150.