For a prime $p$, a natural number $n$ and an integer $a$, we let $S_n(a,p)$ denote the exponent of $p$ in the prime factorisation of $a^{p^n} - 1$.  For example, $S_1(4,3) = 2$ and $S_2(6,2) = 0$.  Find all pairs $(n,p)$ such that $S_n(2013,p) = 100$.
We are given that \( S_n(2013, p) = 100 \), which means the highest power of \( p \) dividing \( 2013^{p^n} - 1 \) is 100. We need to find all pairs \((n, p)\) that satisfy this condition.

1. **Check if \( p \) divides 2013:**
   Since \( p^{100} \mid 2013^{p^n} - 1 \), it implies that \( p \) does not divide 2013 (otherwise, \( 2013^{p^n} - 1 \) would not be divisible by \( p^{100} \)). Therefore, \( \gcd(p, 2013) = 1 \).

2. **Using Fermat's Little Theorem:**
   Fermat's Little Theorem states that for a prime \( p \) and an integer \( a \) such that \( \gcd(a, p) = 1 \), we have \( a^{p-1} \equiv 1 \pmod{p} \). Applying this to our problem, we get:
   \[
   2013^{p-1} \equiv 1 \pmod{p}
   \]
   This implies that \( p \mid 2013^{p-1} - 1 \).

3. **Using Lifting The Exponent Lemma (LTE):**
   The LTE lemma states that for odd primes \( p \), if \( x \equiv y \not\equiv 0 \pmod{p} \), then:
   \[
   v_p(x^n - y^n) = v_p(x - y) + v_p(n)
   \]
   For \( p = 2 \), we have a different form:
   \[
   v_2(x^n - y^n) = v_2(x - y) + v_2(x + y) + v_2(n) - 1
   \]

4. **Case \( p = 2 \):**
   \[
   2013^{2^n} - 1
   \]
   Since \( 2013 \equiv 1 \pmod{2} \), we have:
   \[
   v_2(2013^{2^n} - 1) = v_2(2012) + n = 2 + n
   \]
   Given \( v_2(2013^{2^n} - 1) = 100 \), we get:
   \[
   2 + n = 100 \implies n = 98
   \]

5. **Case \( p \neq 2 \):**
   For odd primes \( p \), we use the LTE lemma:
   \[
   v_p(2013^{p^n} - 1) = v_p(2013 - 1) + v_p(p^n) = v_p(2012) + n
   \]
   Since \( 2013 \equiv 1 \pmod{p} \), we have:
   \[
   v_p(2013 - 1) = v_p(2012)
   \]
   Given \( v_p(2013^{p^n} - 1) = 100 \), we get:
   \[
   v_p(2012) + n = 100
   \]
   We need to find \( p \) such that \( v_p(2012) \neq 0 \). The prime factorization of 2012 is:
   \[
   2012 = 2^2 \times 503
   \]
   Therefore, \( v_{503}(2012) = 1 \). Thus:
   \[
   1 + n = 100 \implies n = 99
   \]

Conclusion:
The pairs \((n, p)\) that satisfy the given condition are \((98, 2)\) and \((99, 503)\).

The final answer is \( \boxed{ (98, 2) } \) and \((99, 503)\).