One of Euler's conjectures was disproved in the $1980$s by three American Mathematicians when they showed that there is a positive integer $n$ such that \[n^{5}= 133^{5}+110^{5}+84^{5}+27^{5}.\] Find the value of $n$.
1. **Divisibility by 2:**
   Since 110 and 84 are even, and 27 and 133 are odd, the sum of their fifth powers will be even. Therefore, \( n^5 \) is even, implying \( n \) is even. Thus, \( 2 \mid n \).

2. **Divisibility by 3:**
   Both 84 and 27 are divisible by 3. We need to check if \( 133^5 + 110^5 \) is divisible by 3. Since \( 133 \equiv 1 \mod 3 \) and \( 110 \equiv 2 \mod 3 \), we have:
   \[
   133^5 \equiv 1^5 \equiv 1 \mod 3
   \]
   \[
   110^5 \equiv 2^5 \equiv 32 \equiv 2 \mod 3
   \]
   Therefore,
   \[
   133^5 + 110^5 \equiv 1 + 2 \equiv 0 \mod 3
   \]
   Since \( 84^5 \equiv 0 \mod 3 \) and \( 27^5 \equiv 0 \mod 3 \), the sum \( 133^5 + 110^5 + 84^5 + 27^5 \equiv 0 \mod 3 \). Thus, \( 3 \mid n \).

3. **Congruence modulo 5:**
   By Fermat's Little Theorem, \( n^5 \equiv n \mod 5 \). We need to find \( n \mod 5 \) such that:
   \[
   133^5 + 110^5 + 84^5 + 27^5 \equiv n^5 \mod 5
   \]
   Calculating each term modulo 5:
   \[
   133 \equiv 3 \mod 5 \implies 133^5 \equiv 3^5 \equiv 243 \equiv 3 \mod 5
   \]
   \[
   110 \equiv 0 \mod 5 \implies 110^5 \equiv 0 \mod 5
   \]
   \[
   84 \equiv 4 \mod 5 \implies 84^5 \equiv 4^5 \equiv 1024 \equiv 4 \mod 5
   \]
   \[
   27 \equiv 2 \mod 5 \implies 27^5 \equiv 2^5 \equiv 32 \equiv 2 \mod 5
   \]
   Summing these:
   \[
   133^5 + 110^5 + 84^5 + 27^5 \equiv 3 + 0 + 4 + 2 \equiv 9 \equiv 4 \mod 5
   \]
   Thus, \( n \equiv 4 \mod 5 \).

4. **Congruence modulo 7:**
   We need to find \( n \mod 7 \) such that:
   \[
   133^5 + 110^5 + 84^5 + 27^5 \equiv n^5 \mod 7
   \]
   Calculating each term modulo 7:
   \[
   133 \equiv 0 \mod 7 \implies 133^5 \equiv 0 \mod 7
   \]
   \[
   110 \equiv 5 \mod 7 \implies 110^5 \equiv 5^5 \equiv 3125 \equiv 3 \mod 7
   \]
   \[
   84 \equiv 0 \mod 7 \implies 84^5 \equiv 0 \mod 7
   \]
   \[
   27 \equiv 6 \mod 7 \implies 27^5 \equiv 6^5 \equiv 7776 \equiv -1 \mod 7
   \]
   Summing these:
   \[
   133^5 + 110^5 + 84^5 + 27^5 \equiv 0 + 3 + 0 - 1 \equiv 2 \mod 7
   \]
   Thus, \( n \equiv 4 \mod 7 \).

5. **Chinese Remainder Theorem:**
   We have:
   \[
   n \equiv 4 \mod 5
   \]
   \[
   n \equiv 4 \mod 7
   \]
   Since 5 and 7 are coprime, by the Chinese Remainder Theorem, we can combine these congruences:
   \[
   n \equiv 4 \mod 35
   \]

6. **Combining with divisibility by 2 and 3:**
   Since \( n \) is divisible by 2 and 3, and \( n \equiv 4 \mod 35 \), we need to find \( n \) such that:
   \[
   n = 35k + 4
   \]
   and \( n \) is divisible by 6. Testing \( k = 4 \):
   \[
   n = 35 \cdot 4 + 4 = 140 + 4 = 144
   \]
   \( 144 \) is divisible by 6, so \( n = 144 \).

The final answer is \( \boxed{144} \).